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Statement

For integers $L_0,L_1$, write $S(L_0,L_1)=(L_0,L_1,L_2,\dots)$ for the sequence with $L_{n+2}=L_{n+1}+L_n$ for $n=0,1,2,\dots$ (p. 322). For a prime $p$, $r(p)$ is its rank of apparition in the Fibonacci numbers $F_n$: the least positive $r$ with $F_r\equiv0\pmod p$ (p. 323).

Main result (unnumbered; announced p. 322, construction p. 323, numbers p. 324). The note sets out to exhibit integers $M$ and $N$ with the following properties (p. 322):

"1. $M$ and $N$ are relatively prime. 2. No term of $S(M, N)$ is a prime number."

The construction (p. 323) takes eighteen primes $a_1,\dots,a_{18}$ with residues $b_n$, listed as $(a_n,r(a_n),b_n)$: $(2,3,2)$, $(3,4,1)$, $(5,5,1)$, $(7,8,3)$, $(17,9,4)$, $(11,10,2)$, $(61,15,3)$, $(47,16,7)$, $(19,18,10)$, $(41,20,10)$, $(53,27,16)$, $(109,27,7)$, $(31,30,24)$, $(2207,32,15)$, $(5779,54,52)$, $(2521,60,60)$, $(1087,64,31)$, $(4481,64,63)$. The paper asserts that the progressions $A_n=\{r(a_n)k+b_n:k\in\mathbb Z\}$, $n=1,\dots,18$, cover the integers, and that any $L_0,L_1$ with

$$L_0\equiv F_{r(a_n)-b_n},\qquad L_1\equiv F_{r(a_n)-b_n+1}\pmod{a_n} \qquad(n=1,\dots,18)$$

(its system (2)) give $L_{b_n}\equiv0\pmod{a_n}$ for every $n$, so that every term of $S(L_0,L_1)$ is divisible by some $a_n$. With John Brillhart's help, the smallest positive solution of (2) is stated to be (p. 324)

$$M=L_0=1786772701928802632268715130455793,\qquad N=L_1=1059683225053915111058165141686995,$$

and the paper concludes that every term of $S(M,N)$ is composite and that $(M,N)=1$ by the Euclidean algorithm.

Correction. A computation made while filing, not filed as evidence, confirms that the eighteen $a_n$ are primes with the printed ranks of apparition, that the eighteen progressions cover the integers (their moduli have least common multiple $8640$), and that $(M,N)=1$. It also finds that the printed pair satisfies (2) for seventeen of the eighteen primes but not for $a_{17}=1087$: there $M\equiv1048$ and $N\equiv524\pmod{1087}$, while (2) requires $F_{33}\equiv524$ and $F_{34}\equiv485$. So $L_{31}\not\equiv0\pmod{1087}$, and the printed numbers are not a solution of (2). The conclusion that every term of $S(M,N)$ is composite is therefore not established by the printed argument for the printed pair; whether it holds for that pair was not settled here. For a pair that does solve (2), the printed argument gives every term a divisor among the $a_n$.

Source. R. L. Graham, A Fibonacci-like sequence of composite numbers, Math. Mag. 37 (1964), no. 5, 322--324; the aim on p. 322, the table, covering and system (2) on p. 323, the numbers and conclusion on p. 324. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the table, system (2) and the two printed integers were read on the printed pages and checked by the computation described above; nothing here is independently reviewed.

Proof pointer

Pages 322--323. The identity $L_{m+n}=F_{n-1}L_m+F_nL_{m+1}$ (the note's (1), by induction on $n$) gives $L_{m+r(p)}\equiv F_{r(p)-1}L_m\pmod p$, so a zero of $L$ modulo $p$ recurs with period $r(p)$. The covering is checked in four steps: six progressions cover the odd integers, six more the remaining integers not divisible by $6$, four more those not divisible by $30$, and the last two the multiples of $30$. Under (2), $L_m\equiv F_{r(a_n)-b_n+m}\pmod{a_n}$, so $L_{b_n}\equiv F_{r(a_n)}\equiv0$; the Chinese remainder theorem gives a simultaneous solution since the $a_n$ are distinct primes.

Bears on

• Problem 276: for a pair that solves (2), each term is divisible by one of the eighteen primes, so their product has a common factor with every term; such a sequence fails the problem's second condition and is not an example for it. The printed pair does not solve (2) (see the Correction), and the paper says nothing about sequences with no such common-factor integer.