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This page follows the two mechanisms of the reconstructed proof of Theorem 1.2 of Pollack, Pomerance and Treviño into the rest of the corpus and asks, for each candidate target, what they give when transported, what they do not give, and what the next investigation is. The first mechanism is the pair of uniform collision bounds: the sketch-level Theorem 3.1 for the unstructured solutions of $\varphi(n)=\varphi(n+k)$ and the reconstructed Theorem 3.3 for the structured ones, with the $S$-unit count of Lemma 3.2 behind it. The second is the convenient-integer construction of Lemma 4.1, which multiplies one totient with known preimages into $\gg Z(x)$ totients with known preimages, and its use through reversed pairs in Lemma 5.1 and the proof of Theorem 1.2.

Standing. Every statement proved on this page is author-recorded, changing no problem status. The statements were independently reviewed as they stood on 2026-09-28T13:50:48Z by eleven focused reviews, one per section, graded in the connections grade, which passes all eleven and accepts twelve corrections, C1--C12, since applied; it assigns no tier. Since the review, the Target paragraphs were updated to the standing their problem pages record, the Problem 1003 one restated without a date, and the first next investigation of the Problem 1004 section was reworded to match; this touches no examined statement, proof, code or input. Each statement is deduced from the source theorems as published, at the scope the reconstruction pages record, and each such input is named where it is consumed. The candidate connections were assessed on the mathematics, by checking hypotheses, quantifier order, uniformity and the actual objects, not on shared vocabulary; connections that fail are kept with their obstruction.

Outcomes at a glance. Problem 1004: transfers for every exponent below 2 (Proposition 1), open at and above 2. Problem 1003: fails. Problem 417: transfers (bounded ratio, Proposition 2; limit superior above 1, Proposition 3) and transfers conditionally (limit inferior above 1, Proposition 4). Problem 416: fails. Problem 694: transfers (Proposition 5), but the target is solved. Problems 51, 821, 415, 850 and 931: fail. The open clause of Problem 49: nothing transfers. The Problem 1004 route is carried by the new lead distinct totient blocks from collision bounds; the Problem 417 route is carried by the lead distinct totient preimage ratio, written in parallel from the Problem 416 reconstruction, whose statements this page reaches independently from this folder's own inputs.

Notation and the inputs used

$\varphi$ is Euler's function, a totient is a value of $\varphi$, and the preimages of a totient $v$ are the integers $m$ with $\varphi(m)=v$. For real $x$ let

$$V(x)=\#\{v\le x:v\text{ a totient}\},\qquad \mathcal W(x)=\{\varphi(n):n\le x\},\qquad W(x)=\#\mathcal W(x),$$

so that $W(x)\le V(x)$; Problem 417 writes $V'(x)$ for $W(x)$. For a natural number $k$, $P(x;k)=\#\{n\le x:\varphi(n)=\varphi(n+k)\}$ splits as $P_0(x;k)+P_1(x;k)$, the structured solutions of Theorem A and the rest, as on the Theorem 3.3 page, and $c(k)$ is the coefficient defined there. $Z(x)$ is Ford's function, defined on the Theorem 1.2 page. An integer $n$ is convenient for a totient $d$ with preimages $n_1,\ldots,n_k$ when $d\varphi(n)$ is a totient whose preimages are exactly $n_1n,\ldots,n_kn$ (Lemma 5.1 page).

The inputs consumed below, at their actual scope:

• (T1) Theorem 3.1 of the source: there is an absolute $x_0$ such that $P_1(x;k)<x/\exp((\log x)^{1/3})$ for $x>x_0$, uniformly for natural numbers $k\le\exp((\log x)^{1/3})$. A published theorem whose proof the source gives as a sketch; the reconstruction records the sketch and its one written deduction only.
• (T3) Theorem 3.3 of the source: for even $k$ with $2\le k\le x^{\varepsilon(x)}$, where $\varepsilon(x)\to0$ and $x^{\varepsilon(x)}\to\infty$, $P_0(x;k)\le(16C_2+o(1))c(k)x/(\log x)^2$ uniformly in $k$, where $C_2=2\prod_{p>2}(1-(p-1)^{-2})$ is the twin-prime constant in the source's normalization (Theorem 3.3 page), and $\frac1{2k}\le c(k)\le\Bigl(3\cdot7^{3+2\omega(k)}\prod_{p\mid k,\ p>2}\frac{p-1}{p-2}\Bigr)\frac1k$. Reconstructed modulo Selberg's sieve as the source states it and Evertse's bound inside Lemma 3.2. For odd $k$, $P_0(x;k)=0$ (Theorem 1.2 page, step 2).
• (L4) Lemma 4.1 of the source: for fixed totients $d_1,d_2$ and fixed $D\ge\max\{d_1,d_2\}$ there are an absolute $K\ge1$, a constant $c_D>0$ and $x_0(D)$ such that for $x\ge x_0(D)$ at least $c_DZ(x)$ integers $n$ satisfy $\varphi(n)\le x/D$, $n$ convenient for $d_1$ and for $d_2$, and $n/\varphi(n)\le K$. A published lemma whose two counting steps import Ford's lower-bound argument; the reconstruction is partial.
• (F) Ford's Theorem 1, on the card Ford (1998): $V(x)=Z(x)e^{O(1)}$ for large $x$. Three consequences are used: there are absolute $c_0,C_1>0$ with $c_0Z(x)\le V(x)\le C_1Z(x)$ for large $x$; and for each fixed $\mu\ge1$, $Z(x/\mu)/Z(x)\to1/\mu$, because $\log(x/\mu)\sim\log x$ and $\log_3(x/\mu)-\log_3x=O(1/(\log x\log_2x))$, so the exponent of $Z$, whose leading term is $C(\log_3x-\log_4x)^2$, changes by $O(\log_3x/(\log x\log_2x))\to0$; hence $V(x/\mu)\le(C_3/\mu)V(x)$ for $x\ge x_2(\mu)$ with an absolute $C_3$.
• (P) The explicit totient $d_1=2^{18}\cdot257=67371008$, whose eight preimages are enumerated on the Lemma 5.1 page; all eight are even. The enumeration was repeated for this page by the same divisor recursion, with the same result; both runs are author-recorded and not filed as evidence.

Problem 1004: distinct totients on a consecutive block

Target. Problem 1004 asks: for fixed $c>0$ and all large $x$, is there $n\le x$ such that $\varphi(n+1),\ldots,\varphi(n+\lfloor(\log x)^c\rfloor)$ are pairwise distinct? Its page records the status open for every $c$. Its one claim, the anonymous note's pending partial claim, asserts the almost-all form for every $c<2$, and the note is also kept as the unreviewed candidate totient_blocks_native_note. Its current assessment records that the cited sources state no growing-block existence theorem, and its Progress records the uniform bounds (T1) and (T3) as the relevant inputs and this page's deduction for $c<2$, Proposition 1 below.

What transports. The union bound over the $\binom L2$ shifts of a block of length $L$, fed with (T1), (T3) and the upper bound on $c(k)$, gives the following. The proof is complete relative to (T1) and (T3).

Proposition 1 (author-recorded). Let $L=L(x)$ be integer-valued with $2\le L(x)\le(\log x)^2/(\log\log x)^{50}$. Then the number of integers $1\le n\le x$ for which $\varphi(n+1),\ldots,\varphi(n+L)$ are not pairwise distinct is $o(x)$. In particular, for every fixed $c<2$ and all large $x$, all but $o(x)$ of the $n\le x$ have $\varphi(n+1),\ldots,\varphi(n+\lfloor(\log x)^c\rfloor)$ pairwise distinct, so Problem 1004 has an affirmative answer for every $c<2$.

Proof. Let $x$ be large, $L=L(x)$, and $y=x+L\le2x$. Let $B(x,L)$ count the integers $1\le n\le x$ with $\varphi(n+i)=\varphi(n+j)$ for some $1\le i<j\le L$.

Step 1: reduction to shifted collisions. If $n$ is counted, put $m=n+i$ and $h=j-i$; then $1\le h\le L-1$, $m\le x+L-h\le y$ and $\varphi(m)=\varphi(m+h)$, so $m$ is counted by $P(y;h)$. Given $(m,h)$, $n=m-i$ with $1\le i\le L-h$, at most $L-h$ values. Hence

$$B(x,L)\le\sum_{h=1}^{L-1}(L-h)P(y;h)\le L\sum_{h=1}^{L-1}P(y;h).$$

Step 2: the uniform inputs. For large $x$, $L\le(\log x)^2\le\exp((\log y)^{1/3})$, so (T1) gives $P_1(y;h)<y\exp(-(\log y)^{1/3})$ for every $1\le h\le L-1$. Take $\varepsilon(y)=(\log y)^{-1/2}$: then $\varepsilon(y)\to0$, $y^{\varepsilon(y)}=\exp(\sqrt{\log y})\to\infty$, and $L\le(\log x)^2\le y^{\varepsilon(y)}$ for large $x$, so (T3) gives, for every even $h\le L-1$ and large $x$, $P_0(y;h)\le(16C_2+1)c(h)\,y/(\log y)^2$, the $o(1)$ being uniform in $h$. For odd $h$, $P_0(y;h)=0$. With $S(L)=\sum_{h\le L,\ h\text{ even}}c(h)$,

$$\sum_{h=1}^{L-1}P(y;h)\le L\,y\exp\bigl(-(\log y)^{1/3}\bigr)+(16C_2+1)S(L)\frac{y}{(\log y)^2}.$$

Step 3: the average of $c(h)$. By the upper bound in (T3), $c(h)\le3\cdot7^3\,g(h)/h$ with $g(h)=49^{\omega(h)}\prod_{p\mid h,\ p>2}\frac{p-1}{p-2}$, a nonnegative multiplicative function with $g(2^a)=49$ and $g(p^a)=49\frac{p-1}{p-2}$ for $p>2$ and all $a\ge1$. For $L\ge3$,

$$\sum_{h\le L}\frac{g(h)}{h} \le\prod_{p\le L}\Bigl(1+\sum_{a\ge1}\frac{g(p^a)}{p^a}\Bigr) =\prod_{p\le L}\Bigl(1+\frac{g(p)}{p-1}\Bigr) =50\prod_{3\le p\le L}\Bigl(1+\frac{49}{p-2}\Bigr) \le50\exp\Bigl(49\sum_{3\le p\le L}\frac1{p-2}\Bigr),$$

and $\sum_{3\le p\le L}1/(p-2)\le\sum_{p\le L}1/p+\sum_{p\ge3}2/(p(p-2))=\log\log L+O(1)$ by Mertens' theorem. Hence $S(L)\le A_0(\log L)^{49}$ for $L\ge3$ with an absolute constant $A_0$, and enlarging $A_0$ covers $L=2$, where the sum is $c(2)$.

Step 4: conclusion. Since $y\le2x$ and $\log y\ge\log x$,

$$B(x,L)\le2x\Bigl[(\log x)^4\exp\bigl(-(\log x)^{1/3}\bigr) +(16C_2+1)A_0\frac{L(\log L)^{49}}{(\log x)^2}\Bigr]$$

for large $x$ (using $L\le(\log x)^2$ in the first term). The first bracket term tends to $0$. In the second, $L\le(\log x)^2/(\log\log x)^{50}$ and $\log L\le2\log\log x$ give $L(\log L)^{49}/(\log x)^2\le2^{49}/\log\log x\to0$. So $B(x,L)=o(x)$, and all but $o(x)$ of the $n\le x$ have the block pairwise distinct. For fixed $c<2$, $L=\lfloor(\log x)^c\rfloor$ satisfies the hypothesis for large $x$. $\square$

What it does not give. The range $c\ge2$, for a reason internal to the method. The first-moment bound of Step 1 is fed with the termwise upper bound $P_0(y;h)\le(16C_2+o(1))c(h)y/(\log y)^2$, and (T3) also proves $c(h)\ge1/(2h)$ for even $h$; so whatever average bound for $c(h)$ is used, the bound on $B(x,L)$ obtained in Step 4 is at least of order $L\log L\cdot x/(\log x)^2$, which is not $o(x)$ once $L\log L\gg(\log x)^2$. The anonymous note's sharper average $\sum_{h\le L}c(h)\ll\log L$ would extend Proposition 1 exactly to the $L$ with $L\log L=o((\log x)^2)$, for instance $L\le(\log x)^2/(\log\log x)^{1+\delta}$ for fixed $\delta>0$, and to no $L\ge\kappa(\log x)^2/\log\log x$ with fixed $\kappa>0$. This is a limit of the union bound, not a proved obstruction to the target: it would become one only with a matching lower bound $P_0(x;h)\gg c(h)x/(\log x)^2$, which is a prime-pair statement (both $ar+1$ and $br+1$ prime) and is not available. If that lower bound holds, as the Hardy--Littlewood heuristic predicts, the expected number of collisions inside a block of length $L$ at a random start $n\le x$ is at least of order $L\log L/(\log x)^2$. A first moment does not decide how many blocks are collision-free. Under a Poisson model for the positions of collisions, for $c>2$ almost all blocks contain a collision and only atypical starts can answer the question, and the same model predicts a longest collision-free block below $x$ of length $(\log x)^{3+o(1)}$, so under it the every-$c$ form would fail for $c>3$. The model is a heuristic and is stated here as such. An exploratory computation (a sieve of $\varphi$ and a sliding window over $1\le m\le N$; author-recorded, not filed) found the longest block of consecutive integers with pairwise distinct totients below $N=10^5,10^6,10^7$ to have lengths $111$, $194$ and $321$, against $(\log N)^2=133,191,260$ and $(\log N)^3/\log\log N=625,1004,1506$; the range is far too short to separate the exponents, and the observation was made after the heuristic was written, not predicted by it.

Relation to the retained note. The note preserved under totient_blocks_native_note claims the same $c<2$ corollary by the same union bound, with the average $\sum_{h\le L}c(h)\ll\log L$ it asserts on its pp. 2--3. Proposition 1 needs no average estimate beyond the crude $(\log L)^{49}$ of Step 3, which follows from the bound on $c(k)$ that (T3) itself states, so the $c<2$ answer rests on (T1) and (T3) alone. The erdosproblems.com discussion thread for the problem, read on 2026-09-28, carries the note and a reader's remark that the statement is implicit in the source paper; the erdosproblems.com problem page, last edited 12 April 2026, records neither.

Next investigation. (1) Proposition 1, whose only inputs are (T1) and (T3), was reviewed with this page (see Standing), and the problem page records the $c<2$ answer under its Progress; it does not note that the source gives (T1)'s proof as a sketch. (2) For $c\ge2$ the question becomes whether the collision set $\mathcal C_L=\{m:\varphi(m)=\varphi(m+h)\text{ for some }1\le h<L\}$ has gaps of length $L$ below $x$. A positive result for some $c\ge2$ would need more than first moments: either a lower bound for the number of gaps (a second-moment or sieve argument exploiting that structured collisions have the shape $m=jq$, $m+h=(j+h)p$ with $q=br+1$ and $p=ar+1$ prime), or a construction of a start $n$ avoiding every such shape. A negative result for some $c$, showing that every block of length $(\log x)^c$ below $x$ contains a collision, would refute the every-$c$ expectation and needs a lower bound for collisions, hence a prime-pair input. The route is carried by the lead distinct totient blocks from collision bounds.

Outcome. Transfers for every $c<2$; open for $c\ge2$.

Problem 1003: the unit shift

Target. Problem 1003 asks whether $\varphi(n)=\varphi(n+1)$ has infinitely many solutions. The problem page records the question as open, with Li's Corollary 1.5, a preprint result, as the strongest unit-shift upper bound and Ford's infinitude results for even shifts divisible by a fixed large integer.

What transports. Only upper bounds, in the wrong direction. (T1) with $k=1$ gives $P(x;1)=P_1(x;1)<x/\exp((\log x)^{1/3})$ for large $x$, the bound of Erdős, Pomerance and Sárközy recorded on the problem page and weaker than Yamada's and Li's. The structured family of Theorem A is empty for $k=1$: it needs $j$ and $j+1$ with the same prime factors, and consecutive integers are coprime (Theorem 1.2 page, step 2). Nothing in the reconstruction constructs a solution.

What it does not give, and why. Infinitude is an existence statement. The reconstructed bounds on $P(x;k)$, (T1) and (T3), are upper bounds; the reconstructed lower bounds, (L4) and Lemma 5.1, count convenient integers and missing totient values, and neither constrains $\varphi(n+1)$ for any $n$ it counts. The one construction in the source chain, Theorem A, produces solutions of $\varphi(n)=\varphi(n+k)$ only from pairs $j,j+k$ with equal radicals, which forces $k$ even; this is the structural reason the even-shift results of Ford (2020) do not reach $k=1$, and the reconstruction adds no way around it.

Next investigation. None from this mechanism. Outcome. Fails: the bounds on $P(x;k)$ are upper bounds, the lower-bound inputs count other objects, and the only constructive ingredient needs an even shift.

Problem 417: totients below x against totients of integers below x

Target. Problem 417 asks whether $\lim V(x)/V'(x)$ exists and whether it is $>1$, where $V'(x)=W(x)$ counts the distinct values $\varphi(m)$, $1\le m\le x$, and $V(x)$ counts the totients $\le x$. Its page records the status open with no claim page; its Current assessment records Theorem 2.2 of the OpenAI release, which bears on the problem without claiming it, and its Progress records the results of the Problem 416 folder's connections page. The erdosproblems.com page, read on 2026-09-28, records that Erdős suggested in 1998 that the limit may be infinite, and its discussion thread is undecided about whether the ratio is bounded.

What transports, unconditionally. Three statements.

Proposition 2 (author-recorded; the boundedness that the source quotes from Ford on its p. 7, rederived from (L4)). There is an absolute constant $C$ with $V(x)\le C\,V'(x)$ for all large $x$; so $\limsup_{x\to\infty}V(x)/V'(x)<\infty$.

Proof. The totient $1$ has preimages exactly $1$ and $2$. Apply (L4) with $d_1=d_2=1$ and $D=K$ (allowed, as $K\ge1=\max\{d_1,d_2\}$): for $x\ge x_0(K)$ there are at least $c_KZ(x)$ integers $n$ with $\varphi(n)\le x/K$, $n$ convenient for $1$, and $n/\varphi(n)\le K$. For each such $n$, $n\le K\varphi(n)\le x$, so $\varphi(n)\in\mathcal W(x)$; and convenience for $1$ says that $\varphi(n)$ is a totient whose preimages are exactly $n$ and $2n$, so $n$ is its least preimage and $n\mapsto\varphi(n)$ is injective on this set. Hence $W(x)\ge c_KZ(x)$, and with (F), $V(x)\le C_1Z(x)\le(C_1/c_K)W(x)$. $\square$

Since $V$ and $W$ are nondecreasing integer-valued functions that increase by at most $1$ across any interval of length at most $1$, and both exceed $c_0x/\log x$ by Chebyshev's bound applied to the values $p-1$ of the primes $p\le x$, the ratio $R(x)=V(x)/W(x)$ satisfies $|R(x+h)-R(x)|\le(1+C)/W(x)\to0$ for $0<h\le1$; a bounded function whose unit-step increments tend to $0$ has every value between its limit inferior and limit superior as a limit point. So the set of limit points of $V(x)/V'(x)$ is a closed interval $[\ell_1,\ell_2]$ with $1\le\ell_1\le\ell_2\le C$, exactly as the cluster-interval argument reconstructed on the Problem 416 research page gives for $V(cx)/V(x)$; the Problem 416 folder index already notes that the argument applies once the ratio is bounded. Erdős's suggestion of an infinite limit, as erdosproblems.com reports it, is excluded by Proposition 2.

Proposition 3 (author-recorded). $\limsup_{x\to\infty}V(x)/V'(x)>1$. More precisely, there are absolute constants $c'>0$ and $J\ge1$ such that for every large $x$ some $y\in[2^{-J}x,x]$ has $V(y)-V'(y)\ge c'V(y)$.

Proof. Let $d=d_1=2^{18}\cdot257$ from (P), with preimages $n_1<\cdots<n_8$, all even. Let $E(t)$ count the totients $v\le t$ all of whose preimages are even.

All-even totients are a positive proportion. Apply (L4) with $d_1=d_2=d$ and $D=d$: for $t\ge x_0(d)$ there are at least $c_dZ(t)$ integers $n$ with $\varphi(n)\le t/d$, convenient for $d$. For each, $v=d\varphi(n)\le t$ is a totient whose preimages are exactly $n_1n,\ldots,n_8n$, all even; and $n\mapsto d\varphi(n)$ is injective, since $n_1n$ is the least preimage of $d\varphi(n)$. Hence $E(t)\ge c_dZ(t)\ge cV(t)$ for $t\ge t_1$, with $c=c_d/C_1$ by (F).

All-even totients in the top half have no preimage below $y$. If $m$ is even then $\varphi(m)\le m/2$, so every preimage of an all-even totient $v$ is at least $2v$. If $y/2<v\le y$, every preimage of $v$ exceeds $y$, so $v\notin\mathcal W(y)$ while $v$ is counted by $V(y)$. Hence, for every $y$,

$$V(y)-V'(y)\ge E(y)-E(y/2).$$

A dyadic scale where the increment is large. By (F) with $\mu=2^{J+1}$, $V(x/2^{J+1})\le C_32^{-J-1}V(x)$ for large $x$; fix $J$ with $C_32^{-J-1}\le c/2$. For $x$ so large that $x/2^{J+1}\ge t_1$,

$$\sum_{j=0}^{J}\bigl(E(x/2^j)-E(x/2^{j+1})\bigr)=E(x)-E(x/2^{J+1}) \ge cV(x)-V(x/2^{J+1})\ge\tfrac c2V(x),$$

so some $j\in\{0,\ldots,J\}$ has $E(x/2^j)-E(x/2^{j+1})\ge cV(x)/(2(J+1))\ge cV(x/2^j)/(2(J+1))$. With $y=x/2^j$ and $c'=c/(2(J+1))$, the displayed inequality gives $V(y)-V'(y)\ge c'V(y)$, that is, $V(y)/V'(y)\ge1/(1-c')$. Since $y\to\infty$ with $x$, $\limsup V(x)/V'(x)\ge1/(1-c')>1$. $\square$

So $\ell_2>1$: if the limit exists, it exceeds $1$. This answers the second question of Problem 417 conditionally on the first; the first (existence) is untouched. The search for small seeds is worth recording: a sieve over $m\le6\cdot10^6$ (author-recorded, not filed) found no totient $\le10^6$ without an odd preimage, so all-even totients are rare among small totients, and the source's $d_1$ is the seed used.

What transports conditionally. The natural strengthening is a limit inferior above $1$, and it follows from one window-form premise.

Premise (W) (not proved here). For every totient $d$ and every real $\lambda>1$ there are $c(d,\lambda)>0$ and $t_0(d,\lambda)$ such that for $t\ge t_0$ at least $c(d,\lambda)Z(t)$ integers $n$ are convenient for $d$ and satisfy $t/\lambda<\varphi(n)\le t$.

(L4) is the case of a one-sided range $\varphi(n)\le t$; (W) asks for the same count in a multiplicative window. It cannot be deduced from (L4) as stated, because (L4) gives a lower bound with an unspecified constant $c_D$ and the window count is that lower bound minus a count at the smaller scale that is only bounded above (by $V(t/\lambda)\asymp V(t)/\lambda$), so no fixed $\lambda$ is forced to work. Ford's Theorem 4, on p. 3 of the paper held by the Ford card and not among the results the card lists (for admissible $\theta$, $y\ge x^\theta$ and a possible multiplicity $k$, $V_k(x+y)-V_k(x)\asymp V(x+y)-V(x)\asymp\frac{y}{x+y}V(x+y)$, stated by Ford to follow from a simple modification of the proofs of Theorems 1 and 2), is the short-interval form of the multiplicity count that the same construction proves; (W) is the intermediate statement that modification produces for convenient integers. It is a premise here, not a result.

Proposition 4 (author-recorded, conditional on (W)). If (W) holds for the totient $d=2$ and $\lambda=3/2$, then $V(x)-V'(x)\gg V(x)$, so $\liminf_{x\to\infty}V(x)/V'(x)>1$.

Proof. The totient $2$ has preimages exactly $3,4,6$. By (W) with $t=x/2$ and $\lambda=3/2$, for large $x$ at least $c(2,3/2)Z(x/2)\ge c''Z(x)$ integers $n$ are convenient for $2$ with $x/3<\varphi(n)\le x/2$ (using (F) for $Z(x/2)\ge Z(x)/3$). For each, $v=2\varphi(n)\le x$ is a totient whose preimages are exactly $3n,4n,6n$; the least, $3n\ge3\varphi(n)>x$, so $v\notin\mathcal W(x)$; and $n\mapsto2\varphi(n)$ is injective. Hence $V(x)-V'(x)\ge c''Z(x)\ge(c''/C_1)V(x)$. $\square$

What it does not give, and why. Existence of the limit: nothing here controls the width $\ell_2-\ell_1$, exactly as the doubling-law folder records for $V(cx)/V(x)$. Proposition 3 cannot be upgraded to a limit inferior without (W), because $E$ could in principle grow in bursts, leaving $E(y)-E(y/2)=o(V(y))$ along a sequence of scales.

Parallel derivation. The same three unconditional statements, and the same conditional one, are proved independently on the connections page of the Problem 416 folder (its Propositions 417.2 and 417.4 and Theorem 417.3), from Ford's Theorem 8 at statement level with the seed $54$, whose preimages $81$ and $162$ are both divisible by $3$, and a geometric sequence of scales; that page carries the lead distinct totient preimage ratio, whose premise (P1), the short-interval form of Ford's Theorem 8, is the premise (W) above in multiplicity form. The proofs here use only the inputs this folder reconstructs, (L4) and the seed (P), so the two pages are independent checks of one another; no second lead is opened for the same target and route.

Next investigation. Prove (W) by carrying Ford's short-interval modification through the construction behind (L4). Two readings of Ford's §5, in the held conversion, make this plausible and are the points to verify: the largest prime $p_0$ of a candidate $n=p_0p_1\cdots p_L$ is counted, for fixed $p_1\cdots p_L$, by the prime number theorem in a range fixed by $\varphi(n)\le x/d$, so restricting $p_0-1$ to a window of ratio $\lambda$ costs a factor depending on $\lambda$ only; and the exclusion of the non-trivial solutions of $d\varphi(n)=\varphi(n_1)$ is an upper bound whose fraction of the candidate set decays in the parameter $M_2$, so it survives restriction to the window once $M_2$ is enlarged. A positive outcome gives $\ell_1>1$ and settles the second question of Problem 417 in the form: the ratio stays above a fixed constant greater than $1$ for all large $x$. A negative outcome, $\ell_1=1$, would mean that along some scales almost every totient $\le x$ has a preimage $\le x$, which Proposition 3 shows cannot hold at all scales. The route is carried by the parallel lead named above.

Outcome. Transfers (Propositions 2 and 3); transfers conditionally (Proposition 4).

Problem 416: an asymptotic formula for the totient count

Target. The second question of Problem 416, an asymptotic formula for $V(x)$. Its page records the standing claimed, through the OpenAI release's pending full claim of an asymptotic equivalent for $V(x)$ whose coefficient has no closed form, with the first question ($V(2x)/V(x)\to2$) answered by two accepted partial claims and Ford's Theorem 1 as the best formula-type result before the release.

What transports. (L4) is the lower-bound half of Ford's order of magnitude; the proof of Proposition 2 above recovers $W(x)\ge c_KZ(x)$ from it and, with the upper bound of (F), $W(x)\asymp V(x)\asymp Z(x)$. Nothing more.

What it does not give, and why. A lower-bound construction gives an order of magnitude, never a formula: the count of convenient integers is $\gg_DZ(x)$ with an unspecified constant, and the families $d\varphi(n)$ built from any finite list of seeds $d$ miss whole classes of totients, as Ford remarks in §7 of the held paper, writing of the belief that an asymptotic formula will require the asymptotics of $V_k(x)/V(x)$ and $V(x;k)/V(x)$ for every $k$. The folder index of the Problem 416 reconstruction states the same gap from the doubling-law side.

Outcome. Fails: the mechanism is an order-of-magnitude construction.

Problem 694: the ratio of the largest to the smallest preimage

Target. Problem 694 asks for the size of $\max_{v\le x}f_{\max}(v)/f_{\min}(v)$ over totients $v$, where $f_{\max},f_{\min}$ are the largest and smallest preimages; the current assessment records it as solved, with $(e^\gamma+o(1))\log\log x$ from a five-page note filed on its library card, whose Proposition 3.1 observes that one collision $\varphi(a)=\varphi(b)$ with $a>b$ gives infinitely many totients with ratio at least $a/b$.

What transports. The convenient-integer construction preserves the whole preimage set up to the common factor $n$, hence preserves this ratio exactly, and does so on a positive proportion of totients.

Proposition 5 (author-recorded). For every totient $d$ with preimages $n_1<\cdots<n_k$, the number of totients $v\le x$ with $f_{\max}(v)/f_{\min}(v)=n_k/n_1$ is at least $c_dZ(x)\gg_dV(x)$ for $x\ge x_0(d)$.

Proof. Apply (L4) with $d_1=d_2=d$ and $D=d$. Each of the $\ge c_dZ(x)$ integers $n$ obtained has $\varphi(n)\le x/d$, so $v=d\varphi(n)\le x$, and $v$ is a totient whose preimages are exactly $n_1n,\ldots,n_kn$; thus $f_{\max}(v)/f_{\min}(v)=n_kn/(n_1n)=n_k/n_1$. Distinct $n$ give distinct $v$, since $n_1n$ is the least preimage of $v$. $\square$

So every ratio that occurs at all occurs on a positive proportion of the totients up to $x$, which strengthens the note's Proposition 3.1 from "infinitely many with ratio at least $a/b$" to "a positive proportion with ratio exactly $f_{\max}(n)/f_{\min}(n)$, hence at least $a/b$", where $n=\varphi(a)=\varphi(b)$. The Problem 416 connections page identifies the note's device with the same multiplication by convenient integers, from the other side.

What it does not give. Anything about the maximum: (L4) fixes the seed $d$ before $x$ grows, so the ratios it produces are bounded, while the extremal ratio grows like $\log\log x$ through seeds that grow with $x$.

Outcome. Transfers, but the target is solved; Proposition 5 is a further result for the problem page, not a route.

Problem 51: totients whose smallest preimage is large

Target. Problem 51 asks for an infinite set of totients $a$ whose least preimage $n_a$ has $n_a/a\to\infty$. Its page records the status open, with a rejected full claim.

What transports. From a seed totient $d$ with least preimage $n_1$, the $\gg_dV(x)$ totients $v=d\varphi(n)\le x$ of Proposition 5 have least preimage $n_1n$ and

$$\frac{n_v}{v}=\frac{n_1n}{d\varphi(n)}=\frac{n_1}{d}\cdot\frac{n}{\varphi(n)} \in\Bigl[\frac{n_1}{d},\,K\frac{n_1}{d}\Bigr]$$

by clause (iii) of (L4). So every attained ratio $n_1/d$ is attained, up to the bounded factor $K$, on a positive proportion of totients.

What it does not give, and why. Growth. The transported ratio is bounded by $Kn_1/d$ for a fixed seed, and a family of seeds $d_j$ with $n_1(d_j)/d_j\to\infty$ is the question itself. The bound $n/\varphi(n)\le K$ is not incidental: it comes from the sizes of the primes in Ford's candidate set (the proof of (iii) on the Lemma 4.1 page), and replacing those primes by small ones to make $n/\varphi(n)$ large would break the convenience count, which is Ford's argument for exactly that set. In the other direction the mechanism gives an upper bound of the wrong kind: the totients it constructs have least-preimage ratio bounded, which is consistent with either answer.

Outcome. Fails: the mechanism transports a ratio within a bounded factor and cannot create unbounded ratios.

Problem 821: totients of very large multiplicity

Target. Problem 821 asks whether, for every $\epsilon>0$, infinitely many $n$ have more than $n^{1-\epsilon}$ preimages. Its page records the standing claimed, through the OpenAI release's pending full claim; the known results it references, Baker and Harman's accepted partial claim and Lichtman's pending one, come from shifted primes without large prime factors.

What transports. For convenient $n$ the multiplicity of $d\varphi(n)$ equals that of $d$, so each attained multiplicity is attained on a positive proportion of totients; this is Ford's Theorem 2, already on the Ford card.

What it does not give, and why. The construction never increases a multiplicity: the preimages of $d\varphi(n)$ are exactly the $k$ products $n_in$. Large multiplicity needs totients with many preimages in the first place, which is a different mechanism (many primes $p$ with $p-1$ dividing a fixed smooth number).

Outcome. Fails.

Problem 415: ordering patterns of consecutive totients

Target. Problem 415 defines $F(n)$ as the largest $k$ such that every one of the $k!$ orderings occurs among $\varphi(m+1),\ldots,\varphi(m+k)$ for some $m$ with $m+k\le n$, and asks whether $F(n)=(c+o(1))\log\log\log n$, whether the decreasing pattern is always the first to fail, and whether the pattern of $\varphi(1),\ldots,\varphi(k)$ is the most frequent. Its page records the standing claimed, with the answer disproved, through Chojecki's pending full claim, which answers all three questions no; Pollack, Pomerance and Treviño's 2012 asymptotic for monotone runs, which answers the first question no, is an accepted partial claim.

What transports. Proposition 1 says that for $c<2$ almost every block of length $(\log x)^c$ carries pairwise distinct totients, hence a strict total order: each such block realizes exactly one of the $L!$ patterns.

What it does not give, and why. The pattern questions live at lengths of order $\log\log\log n$, far below the collision scale, and they ask that every pattern occur; a collision bound says nothing about which pattern a block realizes. Controlling the order of consecutive totients is a construction problem in which small, medium and large primes are handled separately, the method of the source's §8, which is not part of this reconstruction.

Background for the problem page, outside this reconstruction. Theorem 1.5 of the same source, recorded as a statement on the primes card Pollack, Pomerance and Treviño (2013), gives the longest run of consecutive integers in $[1,x]$ on which $\varphi$ is nonincreasing the length $\log_3x/\log_6x+O(\log_3x/(\log_6x)^2)$. A strictly decreasing run is nonincreasing, and $F(n)$ requires the decreasing pattern of length $F(n)$ to occur, so $F(n)\le(1+o(1))\log_3n/\log_6n=o(\log_3n)$: the first question has a negative answer for every $c>0$. The erdosproblems.com page for the problem, read on 2026-09-28 and last edited 28 May 2026, records this consequence of the source paper; the corpus page, accessed 2026-09-04, does not yet.

Outcome. Fails for the pattern questions; the background note is a library-level fact for the page compilers, not a transfer.

Problems 850 and 931: pairs with the same prime factors

Target. Problem 850 asks whether two distinct integers $x,y$ can have $x,y$, then $x+1,y+1$, then $x+2,y+2$ with the same prime factors in each pair; Problem 931 asks whether two disjoint products of at least three consecutive integers can have the same prime factors only finitely often. Both corpus pages are open and not compiled beyond the statement.

What transports. Lemma 3.2 says that for each natural number $k$ at most $3\cdot7^{3+2\omega(k)}$ natural numbers $j$ have $\gamma(j)=\gamma(j+k)$. Take $x<y$, the conditions being symmetric, and $k=y-x\ge1$; for positive $x$ each of the three conditions says that $j=x,x+1,x+2$ is such a $j$. Under the erdosproblems.com wording, over all integers, the same bound holds, because an entry $0$ or $\pm1$ forces $x=y$ or a failing pair, and the lemma's injection $j\mapsto((j+k)/k,-j/k)$ counts every integer $j\notin\{0,-k\}$ with $\gamma(|j|)=\gamma(|j+k|)$. So for each fixed difference $k$ there are at most $3\cdot7^{3+2\omega(k)}$ candidates $x$, boundedly many in terms of $\omega(k)$.

What it does not give, and why. Existence or nonexistence: the union over all $k$ is unbounded, and the lemma counts pairs at one difference without seeing the interaction between the three consecutive conditions, which is the whole content of Problem 850. For Problem 931 the objects are products of blocks, not pairs at a fixed difference, so the lemma's hypothesis is not even met. The tool behind the lemma, Evertse's bound for $u+v=1$ in $S$-units, is a standard input for such problems, but the transported statement adds nothing to it.

Outcome. Fails.

Problem 49: the remaining clause

Problem 49 retains one open clause, whether the primes are a largest subset of $\{1,\ldots,N\}$ with strictly increasing totients. Nothing on this page bears on it. The reconstructed bound $M^\uparrow(x)\le(1-c)W(x)$ exceeds $\pi(x)$ by a factor tending to infinity and is superseded by Tao's rate; the reversed pairs of Lemma 5.1 constrain sets whose totients may repeat, while a strict set already has distinct totients and is bounded by $W(x)$ for the elementary reason on the folder index; and the clause is an exact finite question, which none of the asymptotic mechanisms here addresses. The folder index maps the three clauses.