Question 1030558
If the integer is a square and a cube, then it must be a 6th power of some integer x.


If x is a multiple of 7, then *[tex \large x^6] is a multiple of 7 and is equal to 7k for some integer k. Otherwise, x and 7 are relatively prime, and by Fermat's little theorem, we have *[tex \large x^6 \equiv 1 \pmod{7}], which is equivalent to saying *[tex \large x^6 = 7k+1] for some integer k.