Prove that there is just one solution in integers m > n to 2/p = 1/m + 1/n for p an odd prime.
Solution
We have 2/(2k+1) = 1/(k+1) + 1/((k+1)(2k+1)) for any odd 2k+1, so there is certainly at least one solution.
Suppose 2/p = 1/m + 1/n, then 2mn = p(m+n), so p divides 2mn. But p is odd, so p must divide mn. It is prime, so it must divide m or n. If it divides n, then n ≥ p. But m > n, so m > p. Hence 1/m + 1/n < 1/p + 1/p. Contradiction. So p divides m. Put m = kp. Then 1/n = 2/p - 1/kp = (2k-1)/kp. But k and 2k-1 have no common factor, so 2k-1 must divide p. But p is prime, so p = 2k-1 and n = 1/k. Thus we have a unique solution.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03