Suppose that $ a,b \in \mathbb{Z}$ If 4 divides $ a^2 + b^2$, then $a$ and $b$ are not odd.

Question

I am trying to prove old test exercises and this one is resisting me.

I know you can believe that I meant $a^ 2 + b ^ 2$ is divisible by 4, but no, in the test it says that 4 is divisible by $a ^ 2+ b^2$.

Suppose that $ a,b \in \mathbb{Z}$ If 4 is divisible by $ a^2 + b^2$, then $a$ and $b$ are not odd.

I started like this:

Suppose that a and b are even, then $a = 2m$ and $b = 2n$ for some $m,n \in \mathbb{Z} $.

Thus $$ a^2 = (2m)^2 \ \mbox{and}\ \ b^2= (2n)^2$$ $$ a ^2 + b^2= (2m)^2 + (2n)^2$$ $$ a^2 + b^2 = (4 m^2 + 4n^2) = 2(2m^2+2n^2) $$

Now $$2m ^ 2 + 2n^2\in \mathbb{Z} = k$$ Then

$$ a^2+b^2= 2k$$

Is this a correct proof?/ Any better ways to prove this.

Sorry for the lousy English. Thanks.

Answer 1

I'm going to assume you mean $a^2 + b^2$ is divisible by $4$.

Assume $a, b$ are odd. Then $a^2 + b^2 = (2k +1)^2 + (2j+1)^2 = (4k^2 + 4k + 1) + (4j^2 + 4j + 1) = 2(2k^2 + 2j^2 + 2k + 2j + 1)$.

This is not divisible by $4$ since the highest power of $2$ is in the product is only $1$.

We have created a contradiction, so $a,b$ cannot both be odd and satisfy $a^2 + b^2$ is divisible by $4$.

Answer 2

As Good Morning Captain wrote, I assume you meant that $a^2+b^2$ is divisible by $4$.

Consider the values modulo $4$. If $a$ is even, $a=2n$, then $a^2 = 4n^2$ leaves a remainder of $0$. If $a$ is odd, $a=2n+1$, then $a^2 = 4n^2+4n+1 = 4(n^2+n)+1$ leaves a remainder of $1$.

Since each of $a^2$ and $b^2$ can have remainders of only $0$ or $1$, their sum can only have remainders of $0, 1$, or $2$ and the first ($0$) happens only when both are even.

Answer 3

I did a more thorough investigation and I found this proof.

Prove by contradiction

Assume $4|(a^2+b^2)$, where a and b are both odd.

Then $a=2m+1$, $b=2n+1$

Substitute in the equation and $4|4(m^2+m+n^2+n)+2$

We know that $a|b$ if $a|(b+xa) \forall x \in \mathbb{z}$. Since $(m^2+m+n^2+n)$ is an integer, you have $4|4(m^2+m+n^2+n)+2−4(m^2+m+n^2+n)$ where we have used the above formula with $x = -(m^2 + m + n^2 + n)$. This simplifies to $4|2$, which is a contradiction.

This proof was done by user John Colanduoni in this post Proving by Contradiction

I will proceed to put my post as a duplicate.

Answer 4

I am with the other commenters that they probably meant that $a^2+b^2$ is divisible by $4$, because that would actually make for a somewhat interesting proof, whereas assuming that $4$ is divisible by $a^2+b^2$ really does not. In fact, if $4$ is divisible by $a^2+b^2$, then $a$ and $b$ can both be odd: they can both be $1$, and so $a^2+b^2=1+1-2$, which divides $4$.

OK, but you asked about your own proof: Well, it's pretty bad, to be honest. Your proof as a whole shows that $a^2+b^2$ is even if both $a$ and $b$ are even ... which is not only a trivial result, but most importantly it is not what you need to show: if you were thinking that your proof shows that if $a^2+b^2$ is even then both $a$ and $b$ must be even, you're wrong: $a$ and $b$ both being odd would also result in $a^2+b^2$ being even. And second, why should $a^2+b^2$ be even? The claim that $4$ is divisible by $a^2+b^2$ does not mean that $a^2+b^2$ is even: $1$ divides $4$, but is odd.