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Statement

In the notation of displays (5) and (6) of p. 27 (see the p. 27 conjecture), take $n=2$, so that the product is $(x+1)(x+2)$ and the exponents (6) are the exponents of all primes in its factorization. On p. 28 Erdős writes that for small $n$ all the exponents (6) can of course be distinct, but that he "can not even prove that for $n=2$ there are infinitely many values of $n$ [sic] for which the exponents (6) are all distinct"; the varying quantity is $x$.

He then writes: "No doubt there are infinitely many primes $p$ for which $8p^2+1$ is a prime, thus $\{3,2,1\}$ occurs infinitely often for $n=2$" (p. 28). The intended family is $8p^2\cdot(8p^2+1)=2^3p^2(8p^2+1)$, whose exponents are $3,2,1$ when $p$ is odd and $8p^2+1$ is prime. As printed the suggestion yields a single case: for every prime $p\ne3$ one has $p^2\equiv1\pmod3$, so $3$ divides $8p^2+1$, and $8p^2+1$ is prime only for $p=3$, giving $72\cdot73=2^3\cdot3^2\cdot73$. With $8p^2-1$ in place of $8p^2+1$ the residue argument does not apply. This correction is an observation of this page, not of the paper.

Source. P. Erdős, Miscellaneous problems in number theory, Proceedings of the Eleventh Manitoba Conference on Numerical Mathematics and Computing (Winnipeg, Man., 1981), Congr. Numer. 34 (1982), 25--45; the passage on p. 28. The edition read is identified on the source card.

Read depth. Claims checked: the passage was read clause by clause on the page image. There is no proof to check; the residue computation above is elementary.

Dependencies

None.

Bears on

• Problem 913: the passage poses the problem's question for the product of two consecutive integers and reports that Erdős could not prove it; the family it suggests works only for $p=3$ as printed, and the paper records no result on the problem.