Problem:
Prove that there are infinitely many distinct pairs (a,b) of relatively prime integers a>1 and b>1 such that ab+ba is divisible by a+b.
Solution:
Let n be an odd positive integer, and take a=2nβ1,b=2n+1. Then ab+baβ‘1+3β‘0 (mod4), and ab+baβ‘β1+1β‘0(modn). Therefore a+b=4n divides ab+ba.
Alternate solution: Let p>5 be a prime and let pξ β‘1(mod5). For each such prime p we construct a pair of relatively prime numbers (a,b) that satisfy the conclusion of the problem. Thus, we will get infinitely many distinct pairs (a,b) as required.
Let a=3p+2,b=7pβ2. Then a+b=10p. We have Ο(10p)=4(pβ1)=bβa, where Ο is Euler's function.
Obviously, a and b are odd and not divisible by p. They are not divisible by 5 because pξ β‘1(mod5). Thus, a and b are relatively prime to 10p=a+b, and therefore relatively prime to each other.
Therefore, using Euler's theorem,
ab=aa+Ο(10p)=aaβ
aΟ(10p)β‘aa(mod10p)
and since 10p=a+b,
ab+baβ‘aa+ba(moda+b)
However, since a is odd, aa+ba is divisible by a+b. Hence, ab+ba is divisible by a+b.
The problems on this page are the property of the MAA's American Mathematics Competitions