Use proof by contradiction to prove that if $a$ and $b$ are odd integers, then $4 \nmid (a^2+b^2)$.

Question

I have been working through the following proof:

Use proof by contradiction to prove that if $a$ and $b$ are odd integers, then $4 \nmid (a^2+b^2)$.

Below, I have included screenshots of the solutions manual to my text book. I am unclear about what happens between steps 3 and 4 (below).

Would someone be able to help me clarify what is happening between steps 3 and 4? I don't fully understand why we are moving the $4x$ over to the right hand side of the equation (I checked to see if $4x$ was replacing something, but that doesn't seem to be the case). Furthermore, I do not understand why "$4(\text{integer})+2$" concludes that "$2 \mid 4$". Since the proof shows that the $4x$ has been moved over, what is implied by the left-hand side of the equation? I fully understand the proof up to that point. Additionally, if someone has an alternative suggestion for how to prove this (via contradiction), I would appreciate it.

Answer 1

Clarification of this proof:

I think a 4 is missing before the first is-sign. Then we would get $$4 = 4x-4y^2-4y-4z^2-4z+2$$

Which is consistent with $4x = 4y^2+4y+4z^2+4-2$.

Then the equation is written as $$4 = 4 \times \mathrm{int}+2$$

$$4 = 2 \times 2 \times\mathrm{int}+2$$

$$4 = 2 \times (2 \times\mathrm{int}+1)$$

$$4 = 2 \times a$$

Here we have that $a$ is odd, so $\gcd(4,a)=1$. Hence we must have $4 \mid 2$, contradiction.

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Another way:

Odd numbers are $x \equiv 1,3 \mod 4$

If $x \equiv 1 \mod 4$ then $x^2 \equiv 1 \mod 4$.

If $x \equiv 3 \mod 4$ then $x^2 \equiv 1 \mod 4$.

Thus odd squares are $x^2 \equiv 1 \mod 4$. Thus the sum of two odd squares is 2 mod 4.

Answer 2

Heres an alternate proof, if you are interested:

Any odd number is of the form $4k+1$ or $4k+3$. If you square these, you get $16k^2+4k+1=4(4k^2+k)+1$ or $16k^2+24k+9=4(4k^2+6+2)+1$.

The point is any odd number squared is of the form $4x+1$, where $x$ is an integer. Thus the sum of two odd numbers squared is of the form $4x+1+4y+1=4(x+y)+2$. It follows that this is not divisibly by 4.