3

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?

Joe
  • 143
  • The concept of your proof is fine, but the proof as written suffers from various faults which makes it turn into nonsense at a certain point. For starters, you have two different and clashing uses of $k$: "$a=2k+1$ for some integer $k$" and "$a^2+4=4k$ for $k \ge 1$". You also have two different expressions for the same number: "$a^2+5=4k$" and "$4k-5=(2n+1)^2$" so that $a$ and $2n+1$ are the same number. You'll be very close to a perfectly correct proof if you clean up those issues. – Lee Mosher Oct 11 '14 at 14:18

2 Answers2

1

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.

Pedro
  • 122,002
  • is my solution correct? I am using basic method to solve this question. I know about modulo but at my level i do not want to try it yet. – Joe Oct 11 '14 at 14:14
  • "Your idea is fine." I am just looking at the possiblities $n=4k,4k+1,4k+2,4k+3$. In each case, I can write $n^2$ as $4k$ or $4k+1$, hence the conclusion. – Pedro Oct 11 '14 at 14:22
1

Well, there is a little notation failure. Fisrt, you say that $a=2k+1$ and after, you say that $a^2+5=4k$. Obviously, you don't mean that the fisrt $k$ and the second $k$ are the same number, so you should rename one of them.

Besides that, your proof is correct, but somewhat long.

It is clear that you don't know how, or want to use congruences. But even with no congruences, the proof can be shortened -and clarified, I think.

I'll write a proof of the hard part of the problem, that is, the case in which $a$ is odd.

Since $a$ is odd, there exists some integer $k$ such that $a=2k+1$. Squaring and adding $5$, we get $a^2+5=4k^2+4k+6=4(k^2+k+1)+2$, that is not a multiple of $4$.

I hope it helps. Nevertheless, the best way to learn how to write proofs is writing (and reading) proofs.

ajotatxe
  • 65,084