Problem 907

For the solution, we use the following statement:

Every square number, when divided by 4, leaves a remainder of 0 or 1.

Proof of this statement: If the base of the square number $m^{2}$ is even, i.e., $m=2k$ (with natural $k$), then $m^{2}=(2k)^{2}=4k^{2}$ is divisible by 4 without a remainder. If the base $m$ of the square number is odd, i.e., $m=2k+1$ (with natural $k$), then $m^{2}=(2k+1)^{2}=4k^{2}+4k+1$ is divisible by 4 with a remainder of 1.

Proof of the claim: According to the statement just proved,

$$\begin{aligned} n^{2} & \equiv 0(\bmod 4), \text { or } n^{2} \equiv 1(\bmod 4) \\ 3 n^{2} & \equiv 0(\bmod 4), \text { or } 3 n^{2} \equiv 3(\bmod 4) \\ 3 n^{2}-1 & \equiv 3(\bmod 4), \text { or } 3 n^{2}-1 \equiv 2(\bmod 4) \end{aligned}$$

Thus, the number $3 n^{2}-1$ never leaves a remainder of 0 or 1 when divided by 4 for any integer $n$, and therefore, according to the statement proved above, it cannot be a square number.