14th VMO 1976

Find all integer solutions to m^m+n = n¹², n^m+n = m³.

Answer

(m,n) = (4,2)

Solution

Suppose a prime p divides n. Then from the first equation it must also divide m. So suppose pa is the highest power of p dividing m, and pb is the highest power of p dividing n. Then the first equation gives 12b = a(m+n), and the second gives 3a = b(m+n). Hence 36b = 3a(m+n) = b(m+n)², so m+n = 6 and a = 2b. Hence m = 4, n = 2.

Thanks to Suat Namli

14th VMO 1976

© John Scholes
jscholes@kalva.demon.co.uk
13 February 2004
Last corrected/updated 13 Feb 04