My solution is:
Assume by contradiction that there is at least one number a such that $a^2$ + 5 is divisible by 4. Then a is either odd or even.
Consider the case when a is odd. Then a= 2k+1 for some integer k. Then $a^2$ = $4k^2$ + 4k + 1 so $a^2$ is odd. Then $a^2$ + 5 is an even number. In order for $a^2$ + 5 to be divisible by 4, $a^2$ + 5 = 4p for p ≥ 1, 4p-5 must be a perfect square. Since $a^2$ is odd, therefore 4p-5 must be an odd value thus it is a square of another odd integer.
4p-5 = $(2n+1)^2$ = ($4n^2$ + $4n$ + 1) Thus,$4p$ = $4n^2$ + $4n$ + $6$ => $p$= $n^2$ + $n$ + $1.5$ which shows that p is not an integer therefore 4p-5 cannot be a perfect square. Therefore, $a^2$ + 5 cannot be divisible by 4.
Consider the case when a is even. $a$ = $2k$ for some integer k. Then $a^2$ = $4k^2$. Clearly, $a^2$ is divisible by 4. But $a^2$ + 5 is odd value and clearly it is not divisible by 4.
Thus, in all cases we reach a contradiction. Therefore, $a^2$ + 5 is not divisible by 4 for any integers a.
Am i correct?
Best Answer
Your idea is fine. More compactly, any number is congruent to $0,1,2,3$ modulo $4$. Squaring gives $0,1,0,1$. But $5=-1$ modulo $4$, but it cannot be the case $a^2=-1\mod 4$ for any $a$, by the above.