Two divisors of ( n 2 + 1 ) / 2 summing up to n + 1 par Mohamed

In this short note, we give an affirmative answer to a question of Ayad from [1]. 1. Main result In [1], Mohamed Ayad asked to prove that there does not exist an odd prime p and two positive divisors d1 and d2 of (p2 +1)/2 such that d1 +d2 = p+ 1. In this note, we prove a bit more, namely: Theorem 1.1. There does not exist an odd integer n > 1 and two positive divisors d1 and d2 of (n2 + 1)/2 such that d1 + d2 = n+ 1. The condition n > 1 cannot be dropped since for n = 1 we may take d1 = d2 = 1. Proof. Let n > 1 be an odd integer for which there exist two divisors d1 and d2 as in the statement of the theorem. Since n is odd, we get that n2 ≡ 1 (mod 8), therefore (n2 + 1)/2 is odd. We show that d1 and d2 are coprime. Indeed, if not there exists an odd prime q | gcd(d1, d2). Hence, q | d1 + d2, therefore n ≡ −1 (mod q). Since also q | d1 | n2 + 1, we get that n2 ≡ −1 (mod q). From the above two congruences we obtain that (−1)2 ≡ −1 (mod q), so q | 2, which is impossible. Since d1 and d2 divide (n2 + 1)/2 and are coprime, we get that d1d2 | (n2 + 1)/2. Write d1d2 = (n2 + 1)/(2d). Then (d1 − d2) = (d1 + d2) − 4d1d2 = (n+ 1) − 2 ( n2 + 1 d ) = ((d− 2)n+ d)2 + 4− 4d d(d− 2) . Now notice that all divisors of n2 + 1 are congruent to 1 modulo 4. In particular, this applies to d1, d2 and d. Hence, n = d1 +d2−1 is congruent Manuscrit reçu le 26 septembre 2006. 562 Mohamed Ayad, Florian Luca to 1 modulo 4 as well. It follows that X = ∣∣∣∣(d− 2)n+ d 4 ∣∣∣∣ , Y = ∣∣∣∣d1 − d2 4 ∣∣∣∣ , s = d− 1 4 are non-negative integers satisfying (1.1) X −DY 2 = s, with D = d(d − 2) = (4s)2 − 1. If s = 0, then X2 + Y 2 = 0, so Y = 0, leading to d1 = d2. Since these two divisors are also coprime, we get that d1 = d2 = 1, therefore n + 1 = d1 + d2 = 2, contradicting the fact that n > 1. Hence, we may assume that s ≥ 1. The above Diophantine equation (1.1) leads to ∣∣∣XY −D ∣∣∣∣ = s Y (X +DY ) . Note that X > Y √ D, therefore X + √ DY > 2 √ DY > 2sY , because √ D = √ (4s)2 − 1 > s, therefore ∣∣∣XY −D ∣∣∣∣ < 1 2Y 2 . It is a known criterion due to Legendre that the above inequality implies that X/Y must be a convergent of √ D. With λ = 4s, it is easily checked that we have the following continued fraction expansion √ λ2 − 1 = [λ− 1, 1, {2(λ− 1), 1}], where {. . .} emphasizes the period. Using the above continued fraction, one checks easily that if pm/qm denotes the mth convergent to √ λ2 − 1, then pm − Dq2 m = −2λ + 2 or 1, according to whether m is even or odd. Hence, for our values of X and Y we should have that X −DY 2 ∈ {−8s+ 2, 1}, and comparing it with equation (1.1) we get that the only chance is s = 1, leading to d = 5 and D = 15. Hence, (X,Y ) is a solution of the Pell equation (1.2) X − 15Y 2 = 1. The minimal solution of the above Pell equation is (X1, Y1) = (4, 1). Hence, if we write (Xt, Yt) for the tth solution of Pell equation (1.2), we then get that Xt + √ 15Yt = (4 + √ 15) holds for all t ≥ 1. Using the above representation, one checks easily that Xt ≡ 1 (mod 3) for all positive integers t. Thus, if Xt = s(n+1)−(n−1)/4 = (n+1)−(n−1)/4 for some positive integer n, we would then get that n = (4Xt − 5)/3, but since 4Xt − 5 ≡ −1 (mod 3), we get that (4Xt − 5)/3 is never an integer Two divisors of (n + 1)/2 summing up to n + 1 563 for any positive integer t. Thus, there is no solution and the theorem is completely proved.