No $n$ makes both $(21n-3)/4$ and $(15n+2)/4$ integers

Prove that there is no $n \in \mathbb{N}$ so that $(21n - 3)/4$ and $(15n + 2)/4$ are both integers.

Solution

If $(21n-3)/4$ is an integer, then \[ 21n-3\equiv0\pmod4. \] Since $21\equiv1\pmod4$, this gives \[ n\equiv3\pmod4. \] If $(15n+2)/4$ is an integer, then \[ 15n+2\equiv0\pmod4. \] Since $15\equiv-1\pmod4$, this gives \[ n\equiv2\pmod4. \] No integer can satisfy both congruences. Therefore no such $n$ exists.