35th Eötvös Competition Problems 1931

1.  Prove that there is just one solution in integers m > n to 2/p = 1/m + 1/n for p an odd prime.

2.  Show that an odd square cannot be expressed as the sum of five odd squares.

3.  Find the point P on the line AB which maximizes 1/(AP + AB) + 1/(BP + AB). 

Solutions

Problem 1

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. 

Problem 2

Show that an odd square cannot be expressed as the sum of five odd squares.

Solution

An odd square must be 1 mod 8. So the sum of five odd squares is 5 mod 8 and cannot be a square. 

Problem 3

Find the point P on the line AB which maximizes 1/(AP + AB) + 1/(BP + AB).

Solution

wlog AB = 1. P must lie on the segment AB, for otherwise we can reduce both AP and BP by moving P closer to the segment. So take P on the segment a distance x from A. Then we want to maximize 1/(1+x) + 1/(2-x) = 3/(2+x-x²) subject to 0 ≤ x ≤ 1. But 2+x-x² = 9/4 - (x - 1/2)², so we minimize the quadratic by taking x = 0 or 1. In other words we maximize the required expression by taking P = A or B.