NT-B2-M01-P018 A Square and a Cube

Show that an odd common prime divisor is impossible by working with \(ab^{-1}\).

Let an odd prime \(p\) divide both expressions. Since \(p\nmid b\), put \(t\equiv ab^{-1}\pmod p\). From \(a^2+b^2\equiv0\), we get \(t^2\equiv-1\); from \(a^3+b^3\equiv0\), we get \(t^3\equiv-1\). Dividing the second congruence by the first gives \(t\equiv1\), but then \(1\equiv-1\pmod p\), a contradiction. Thus there are no odd common prime divisors. If one of \(a,b\) is even, both expressions are odd, so the gcd is \(1\). If both are odd, both expressions are even and \(a^2+b^2\equiv2\pmod4\), so the gcd is \(2\).

This is a strong test of whether students can use inverses even when the modulus is not explicitly given.