Question

if x and y are prime to 3, prove that x^2 + y^@ cannot be a perfect square

Answer 1

Let $n = 3m + k, k = \overline{0,2}, \forall n, m \in N$.

Then $n^2 = 9m^2 + 6mk + k^2 = \begin{cases} 9m^2 \\ 9m^2 + 6m + 1 \\ 9m^2 + 12m + 4 \end{cases} = \begin{cases} 3(3m^2) \\ 3(3m^2 + 3m) + 1 \\ 3(3m^2 + 4m + 1) + 1 \end{cases}$, $k = 0$, $k = 1$, $k = 2$.

Hence, $n^2 = 3l + k, k = \overline{0,1}, \forall n, l \in N$. But $n$ is prime to 3, hence, $n^2$ is prime to 3 and in our case $n^2 = 3l + 1, \forall n, l \in N$.

Consider $x^{2} + y^{2} = (3u + 1)^{2} + (3v + 1)^{2} = 9u^{2} + 6u + 1 + 9v^{2} + 6v + 1 = 3(3u^{2} + 3v^{2} + 2u + 2v) + 2$.

Hence, $x^{2} + y^{2} = 3w + 2$. But we proved, that the square of natural number can have remainder equal to 0 or 1 and in our case it equal to 2. Then $x^{2} + y^{2}$ can't be a perfect square.