This is for the case
If r, s, and t are positive integers such that (r,s) = 1 and , then there are integers m and n such that and .
If r = 1 or s = 1, then this is obviously true, so we may suppose that r > 1 and s > 1. Let the prime-power factorizations of r, s, and t be
Because (r,s) = 1, the primes occurring in the factorizations of r and s are distinct. Because , we have
From the fundamental theorem of arithmetic, the prime-powers occurring on the two sides of the above equation are the same. Hence, each must be equal to for some j with matching exponents, so that . Consequently, every exponent is even, and therefore is an integer. We see that and , where m and n are the integers
Maybe this will help make a general case.