Problem 12019
(American Mathematical Monthly, Vol.125, January 2018) Proposed by N. Safaei (Iran).
Find all positive integers n such that(2n− 1)(5n− 1) is a perfect square.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For n = 1, (21− 1)(51− 1) = 22. For n ≥ 2, we claim that (2n− 1)(5n− 1) is not a perfect square.
For n ≥ 1, (2n− 1)(5n− 1) is a perfect square if and only if 2n− 1 = du2and 5n− 1 = dv2with u, v are positive integers and d is positive square-free integer (an integer which is divisible by no perfect square other than 1). We consider four different cases.
1) If n = 4j + 1 with j ≥ 1 then
(2n− 1)(5n− 1) = (2 · 16j− 1)(5 · 625j− 1) ≡ (−1)(5 − 1) ≡ 12 (mod 16) which is not a quadratic residue modulo 16.
2) If n = 4j + 3 with j ≥ 0 then
(2n− 1)(5n− 1) = (8 · 16j− 1)(5n− 1) ≡ (3 − 1)(−1) ≡ 3 (mod 5) which is not a quadratic residue modulo 5.
3) If n = 2j with j ≥ 1 and d = 1 then 2n− 1 = u2 and 5n− 1 = v2imply
1 = 2n− u2= (2j− u)(2j+ u) > 2 and 1 = 5n− v2= (5j− v)(5j+ v) > 5 which do not hold.
4) If n = 2j with j ≥ 1 and d > 1 then 2n− 1 = du2 and 5n− 1 = dv2 the it follows that d, u are odd, and v is even. Note that (x, y) = (2j, u) and (x, y) = (5j, v) are solutions of the Pell’s equation (d is not a perfect square),
x2− dy2= 1.
Let (x1, y1) be the fundamental solution, then any other solution (xk, yk) is such that xk+ yk
√d= (x1+ y1
√d)k
for some positive integer k. We have that x2k+ y2k
√d= (xk+ yk
√d)2= x2k+ dyk2+ 2xkyk
√d= (2x2k− 1) + 2xkyk
√d
which implies that x2k is odd and y2k is even. Moreover x2k+1+ y2k+1
√d= (x2k+ y2k
√d)(x1+ y1
√d)
= (x2kx1+ dy2ky1) + (x2ky1+ y2kx1)√ d
which implies that x2k+1 has the same parity of x1, and y2k+1 has the same parity of y1. Hence (2j, u) = (xi, yi) for some odd integer i and (5j, v) = (xl, yl) for some even integer l = 2k. Finally
5j = x2k= 2x2k− 1 =⇒ x2k ≡ 3 (mod 5) which does not hold because 3 is not a quadratic residue modulo 5.
