Question

What is the least positive integer value of $x$ such that $(2x)^2 + 2\cdot 37\cdot 2x + 37^2$ is a multiple of $47$?

Answer 1

Rewrite the expression as (2x + 37)^2. For this to be ≡ 0 (mod 47) we need 2x + 37 ≡ 0 (mod 47). Thus 2x ≡ -37 ≡ 10 (mod 47). Multiplying by 2^{-1} ≡ 24 gives x ≡ 24·10 ≡ 240 ≡ 5 (mod 47).

The least positive integer x is 5.