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Statement

Lemma 2 (p. 5). For any positive odd integer $m$, the Fibonacci number $F_m$ has no prime factor of the form $4l+3$.

Use in the paper (pp. 5--6). For the construction of Theorem 3, the paper seeks quadruples $(p_i,m_i,r_i,c_i)$ with $p_i$ prime, $p_i\mid F_{m_i}$, the classes $r_i\bmod m_i$ covering every even integer, and $1\le c_i\le p_i-1$, and sets $x_0\equiv c_iF_{m_i-r_i}$, $x_1\equiv c_iF_{m_i-r_i+1}\pmod{p_i}$ (its (8)). Combined with the square condition $x_0^2+x_0x_1-x_1^2=k^2$ behind Theorem 1, this makes $(-1)^{m_i-r_i+1}$ a quadratic residue modulo $p_i$. The paper concludes that every odd $p_i$ is $\equiv1\pmod4$: by Lemma 2 when $m_i$ is odd, and, when $m_i$ is even, because $r_i$ is then even and $-1$ must be a residue.

Source. Dan Ismailescu and Jaesung Son, A New Kind of Fibonacci-Like Sequence of Composite Numbers, J. Integer Seq. 17 (2014), Article 14.8.2; Lemma 2 and its proof on p. 5, its use on pp. 5--6. The edition read is identified on the source card.

Read depth. Claims checked: the statement and its use were read clause by clause on the printed pages; the short proof was read through.

Proof pointer

Page 5. For odd $m$ and a prime $p\mid F_m$, Cassini's identity $F_{m+1}^2-F_mF_{m+2}=(-1)^m$ gives $F_{m+1}^2\equiv-1\pmod p$, so $-1$ is a quadratic residue modulo $p$, which for an odd prime means $p\equiv1\pmod4$; the prime $2$ is not of the form $4l+3$.

Bears on

• Problem 276: a constraint on the paper's construction only. It restricts which primes can serve in the covering of the even-indexed terms, and says nothing about the problem itself.