8th Irish 1995 Problem A2 Find all integers n for which x 2 + nxy + y 2 = 1 has infinitely many distinct integer solutions x, y.
all n except -1,0,1 Solution Note that (x,y) = (1,0) is always a solution. If (x,y) = (a,b) is a solution, then so is (na+b,-a) because (na+b) 2 -n(na+b)a + a 2 = b 2 + nab + a 2 = 1. But if a > |b|, then na+b > na - a > a for n > 2. So we get an infinite sequence of distinct solutions for n > 2. For n = 2, lhs = (x+y) 2 , so we have infinitely many solutions (x,y) = (k+1,-k). If (x,y) is a solution for n, then (x,-y) is a solution for -n. So we also have infinitely many solutions for n ≤ -2. It remains to consider n = -1, 0, 1. n = 0 obviously has no solutions except (±1,0) and (0,±1), since both x 2 and y 2 are non-negative. We show that n = 1 has no solutions with |x| or |y| > 1. Clearly such a solution must have one of x, y negative. wlog x > 0 > y. So x, -y are positive and hence -2xy ≤ x 2 + y 2 . So x 2 + xy + y 2 ≥ (x 2 + y 2 )/2 > 1, since at least one of x 2 , y 2 ≥ 4. Thus n = 1 has only finitely many solutions. Since any solution (x,y) for n = 1 gives a solution (x,-y) for n = -1, there are only finitely many solutions for n = -1 also. 8th Irish 1995 © John Scholes jscholes@kalva.demon.co.uk 1 February 2004 Last updated/corrected 1 Feb 04