How do you prove for all $k$ in the integers, $2-5k$ can never be a perfect square?

Question

Proving this for $k > 0$ is easy and intuitive, but I'm unsure how to proceed with the negative values for $k$. Should I use modular congruence somehow? Or can I prove this more simply? My gut said since it is for all integers, I should try proof by induction, but I have no idea where to go with that.

Answer 1

Note than any number $n$ ends with the ones digits $0,1,2,3,4,5,6,7,8,9$. If we square $n$, we multiply the one digits, and do some carrying to get $0,1,4,5,6,9$ as the one digits of $n^2$. Now let's examine the ones digits of $2-5k$ for various values of $k$. For $k=0$, ones digit is $2$, however a square doesn't end in $2$ as we have shown before. For $k=-1$ we have $2+5 =7$, but a square can't end in $7$. For $k=-2$ we have $2+10 = 12$, which ends in $2$, but we can't have it end in $2$. Notice that every time we increase $k$ by $1$, the ones digit switches from $2$ to $7$ or $7$ to $2$. This is because you are only adding $5$ each time, thus if you add it to $2$ it becomes $7$, and if you add it to $7$ you get $12$ which has units digit $2$. Since we have shown that a series will switch between $2,7$ as the ones digits if it adds $5$ each time and we have shown that the base cases have $2,7$ as units digits, we have proven that $2-5k$ will always end in $2,7$ (assuming it is positive). However, a square cannot end in $2,7$, thus $2-5k$ can never be a square.

Answer 2

For every integer $x$ we have $x\equiv 0,\pm1, \pm2 \pmod 5$ but then we have $x^2\equiv 0,1,-1\pmod 5$. So if $2-5k = x^2$ for some integers $x$ and $k$, we would have $$x^2 \equiv 2\pmod 5$$ A contradiction.

Answer 3

Simply check that, mod $5$, $0^2, 1^2, 2^2, 3^2$, and $4^2$ are respectively $0,1,4,4$, and $1$. Meanwhile $2-5k\equiv 2$.