16th APMC 1993 Problem 1 Find all positive integers m, n such that 2 m - 3 n = 7.
m=4, n=2 Solution 3 n + 7 = 1 mod 3, and 2 m = (-1) m mod 3, so m is even. It is easy to check that there are no solutions with m = 1, 2, so 2 m = 0 mod 8, and so 3 n = 1 mod 8. Hence n is even. So we have (2 m/2 + 3 n/2 )(2 m/2 - 3 n/2 ) = 7. Hence 2 m/2 + 3 n/2 = 1 or 7. It is obviously too big to be 1, so it must be 7. It is now easy to check that m = 4, n = 2 is the only solution. 16th APMC 1993 © John Scholes jscholes@kalva.demon.co.uk 30 July 2002 Last corrected/updated 30 Jul 02