# Proving that $a,b$ are even integers

I'm trying to prove the following theorem:

Let $$a,b\in\mathbb{Z}$$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$

I always struggle to prove some number is odd or even. How to prove it? I thought of using the $$(a+b)^2=a^2+2ab+b^2$$ formula but not sure how.

• There're only four cases to enumerate. – Saad Apr 10 '19 at 9:41

If $$c$$ is even, then $$c^2\equiv0 \pmod 4;$$ if $$c$$ is odd, then $$c^2\equiv1\pmod4$$.

Therefore if $$a$$ and $$b$$ are both even,

then $$a^2+b^2\equiv0\pmod4;$$

if $$a$$ and $$b$$ are both odd,

then $$a^2+b^2\equiv2\pmod4;$$

and if one of $$a$$ and $$b$$ is even and the other is odd,

then $$a^2+b^2\equiv1\pmod4.$$

Hint: If $$a$$ or $$b$$ or both are odd, then $$(2c+1)^2 = 4c^2+4c+1 \equiv 1 \mod 4$$ and $$(2c+1)^2 + (2d+1)^2= 4c^2+4c+4d^2+4d +2 \equiv 2 \mod 4$$.

But if $$a$$ and $$b$$ are even, then $$(2c)^2+(2d)^2 = 4c^2+4d^2\equiv 0\mod 4$$.

• what is $c$? why did you look at $(2c+1)^2$? Also what if only one of the numbers is odd? – vesii Apr 10 '19 at 9:52
• @vesii: If $a$ is odd,then it can be written as $$a=2c+1$$ for some $c \in \Bbb Z$ – Chinnapparaj R Apr 10 '19 at 9:53
• @Wuestenflux You say "if a or b or both are odd..." but only consider the case when they are both odd. – PierreCarre Apr 10 '19 at 19:21

$$\Rightarrow$$:

By the rules of modular arithmetic we have \begin{align} 0 &= (a^2 + b^2) \bmod 4 \\ &= \left( \left( a^2 \bmod 4 \right) + \left( b^2 \bmod 4 \right) \right) \bmod 4 \end{align} For each summand we have \begin{align} x^2 \bmod 4 &= \left( \left( x \bmod 4 \right) \left( x \bmod 4 \right) \right) \bmod 4 \\ &\in \{ y^2 \bmod 4 \mid y \in \{0,1,2,3\} \} = \{ 0, 1 \} \quad (*) \end{align} So for the expression $$\left( \left( a^2 \bmod 4 \right) + \left( b^2 \bmod 4 \right) \right) \bmod 4$$ we have the possible cases $$0+0$$, $$0+1$$, $$1+0$$ and $$1+1$$. This vanishes modulo $$4$$ only for the case $$0+0$$.

Now $$x^2 \bmod 4$$ only vanishes for $$x \bmod 4 \in \{ 0, 2 \}$$, see display $$(*)$$, which means $$x = 4k$$ or $$x = 4k + 2$$ for some integer $$k$$ and both cases can be divided by $$2$$, thus are even.

$$\Leftarrow$$:

$$a = 2p$$ and $$b=2q$$ for some integers $$p, q$$. Then we have \begin{align} (a^2 + b^2) \bmod 4 &= (4p^2 + 4q^2) \bmod 4 \\ &= \left( \left(4p^2 \bmod 4\right) + \left(4q^2 \bmod 4 \right)\right) \bmod 4 \\ &= (0 + 0) \bmod 4 \\ &= 0 \end{align}

If both $$a$$ and $$b$$ are odd numbers then $$a = 2k_1+1, b=2 k_2+1$$ and you have that $$a^2+b^2 = (2k_1+1)^2+(2 k_2+1)^2 = 4k_1^2+4 k_2^2 +4k_1+4k_2+2 \equiv 2 (mod\,\, 4)$$

If one is odd (for instance $$a$$) and the other is even, then $$a =2 k_1+1, b=2k_2$$ and

$$a^2+b^2 = (2k_1+1)^2+(2k_2)^2 = 4k_1^2+4k_2^2+4k_1+1\equiv 1 (mod\,\,4)$$

If both $$a$$ and $$b$$ are even, $$a=2k_ 1, b=2k_2$$ and $$a^2+b^2 = (2k_1)^2+(2k_2)^2=4k_1^2+ 4k_2^2 \equiv 0(mod\,\,4)$$

So you see that $$a^2+b^2$$ can only be congruent with 0,1 or 2 (mod 4) and is congruent with 0 if and only if both $$a$$ and $$b$$ are even numbers.

$$a^{2}+b^{2}\equiv0\pmod 4 \iff(a+b)^2-2ab\equiv 0\pmod{4}.$$ If both $$a$$ and $$b$$ are odd, then: \begin{align}a+b\equiv 0\pmod{2} &\Rightarrow (a+b)^2\equiv 0\pmod{4}, \text{but}\\ 2ab\equiv 0 \pmod{2} &\Rightarrow \qquad 2ab\not\equiv 0\pmod{4}\end{align} If $$a$$ is odd and $$b$$ is even, then: \begin{align}2ab&\equiv 0 \pmod{4}, \text{but}\\ a+b\not\equiv 0\pmod{2} \Rightarrow (a+b)^2&\not\equiv 0\pmod{4}\end{align} Finally, if $$a$$ and $$b$$ are even, then: \begin{align}a+b\equiv 0\pmod{2} \Rightarrow (a+b)^2&\equiv 0\pmod 4\\ 2ab&\equiv 0\pmod 4\end{align}