bibkey: cohn1964squarefibonacci authors: J. H. E. Cohn year: 1964 title: “Square Fibonacci Numbers, Etc.” doi: null url: https://www.fq.math.ca/Scanned/2-2/cohn2.pdf claim: “Theorems 1 and 2 classify Lucas squares and twice squares at every natural index: the exceptional index sets are {1,3} and {0,6}.” strata_touched:

• D5/S3/Arith/Primes/LucasSquareClassification license: citation-only triage: anchor

Lucas square and twice-square classifications

J. H. E. Cohn, Square Fibonacci Numbers, Etc., Fibonacci Quarterly 2(2) (1964), 109–113. The original primary paper gives the complete Lucas square classification in Theorem 1 and the complete twice-square classification in Theorem 2. The Lucas normalization is L_0 = 2, L_1 = 1, L_(n+2) = L_(n+1) + L_n.

For every natural n, L_n is an integer square exactly for n = 1 or 3; L_n equals twice an integer square exactly for n = 0 or 6. The four exceptional values are 1, 4, 2 and 18, respectively.

The proof uses the signed Lucas trace identity, dyadic Lucas moduli equal to three modulo four, and quadratic nonresidues detected by Jacobi symbols. In the twice-square case, the modulus attached to the residue minus thirty-six is coprime to three. The arithmetic includes index zero explicitly.

Classical argument

The exceptional values are $L_0=2$, $L_1=1$, $L_3=4$, and $L_6=18$. For the exclusion argument, extend the trace to signed indices by $L_{-u}=(-1{)}^u{L}_u$. Trace and norm in the golden ring give

$$L_{j+2k}+(-1{)}^k{L}_j=L_k{L}_{j+k}(j\in \mathbb{Z},k\in \mathbb{N}).$$

If $k$ is even and $s$ is positive and odd, repeated reduction gives $L_{j+2ks}\equiv -L_j(\mathrm{mod}{L}_k)$. Lucas doubling gives $L_{2^{r+1}}\equiv 3(\mathrm{mod}4)$ for every $r\ge 0$.

Every positive even index $n=2m$ is excluded from the square case by $L_{2m}=L_m^2-2(-1{)}^m$: its residue modulo four is two or three. For odd $n$ outside $\{1,3\}$, write $n=j+4t$ with $j\in \{1,3\}$ and $t>0$. Factor $t=2^{r}s$ with $s$ odd, and put $k=2^{r+1}$. The preceding congruence makes $L_n$ congruent to $-1$ or $-4$ modulo the positive odd integer $L_k\equiv 3(\mathrm{mod}4)$. Both residues have Jacobi symbol minus one, so neither is a square.

For the twice-square case, Lucas parity gives $L_n$ even exactly when $3\mid n$. At an odd multiple of three, the Lucas recurrence modulo eight gives $L_n\equiv 4(\mathrm{mod}8)$, which is incompatible with $2x^2$. For $n=4t>0$, the same dyadic factorization with $j=0$ makes $2L_n\equiv -4(\mathrm{mod}{L}_k)$. For $n=6+8t$ with $t>0$, use $j=6$ and $k=2^{v_2(t)+2}$ to obtain $2L_n\equiv -36(\mathrm{mod}{L}_k)$. For $n=2+8t$, use $j=-6$ and factor $t+1$ instead. In the latter two cases, $k$ is divisible by four and Lucas doubling modulo three gives $3\nmid L_k$. Hence the Jacobi symbol of $-36$ is minus one. Since $2L_n=4x^2$ would itself be a square, these congruences exclude all the remaining indices.

The repository formalization is lucas_square_classifications in D5/S3/Arith/Primes/LucasSquareClassification.lean. The general Fibonacci and Lucas square-class theorems additionally need product exclusions at distinct indices, as proved in Ribenboim, Square Classes of Fibonacci and Lucas Numbers, Portugaliae Mathematica 46 (1989), 159–175.

Only the bibliographic citation and a paraphrase of the mathematical result are retained; the paper’s PDF is an external source.