### Testing Dujella's unicity conjecture

The conjecture states that the diophantine equation x^{2} - (k^{2} + 1)y^{2} = k^{2} has at most one positive solution (x,y) with 1 ≤ y < k-1. (For background, see slides.)
We test the conjecture for k in the range m ≤ k ≤ n, where n < 10^{5}, exhibiting the k for which an exceptional solution exists.

If (x,y) is an exceptional solution, let d = gcd(x + k, x − k), a = gcd((x + k)/d, k^{2} + 1), b = gcd((x − k)/d, k^{2} + 1), p^{2} = (x+k)/da, q^{2} = (x-k)/db.
Then ab = k^{2} + 1, 2 < a, 2 < b, gcd(a,b) = 1, d divides 2k, (d distinct from k and 2k, and d is even if k is odd) and ap^{2} - bq^{2} = 2k/d with gcd(p,q)=1.

(p,q) is found by considering the initial period of the continued fraction of √(b/a) (we can assume a < b by interchange): if d > 1, then p/q = A_{m}/B_{m} (a convergent), where Q_{m+1} = 2k/d, whereas if d = 1, then p/q = (A_{m} + eA_{m-1})/(B_{m} + eB_{m-1}), (a quasi-convergent) where | Q_{m} - Q_{m+1} + 2eP_{m+1} | =2k.

We verify that the unicity conjecture holds for k, by showing that there is at most one (d,a,b) for which a solution (p,q) exists and in the case of solubility, that we have precisely one solution (p,q) of ap^{2} - bq^{2} = ± 2k/d with gcd(p,q)=1 and satisfying dpq < k-1.

*Last modified 5th February 2015*

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