When is a number even? Why does something like $a^2 = 2b^2$, show that $a^2$ and thus $a$ are even numbers?
My feeling is that it's because one can divide $a^2$ by two and hence it must be even.
Can anybody give me a clear explanation as to why $a$ is even (in this case) more specifically? Is there mathematical proof?
 A: The multiple-of-2 reason has been raised by a number of others.
Interestingly, $a^2 = 2b^2$ has no integer solution other than $a=b=0$.
The form arises from a proof that $\sqrt 2$ is not rational. That is, $\not \exists a,b \in Z : \sqrt 2 = {a \over b}$. Squaring both sides gives you the equation you started with. The proof starts with assuming that $a,b$ are co-prime. As others have observed, $a^2$ is even, hence $a$ is even. So $a^2$ is a multiple of $4$, making $b$ even, contradicting the co-prime assumption.
A: The key observation is that odd $\times$ odd $=$ odd.
Now, $a^2=2b^2$ implies that $a^2$ is even. If $a$ were odd, then $a^2=a\cdot a$ would be odd. Hence $a$ is even.
A: By definition a number $x$ is even if there exists an integer $y$ such that $x = 2y$. For your case you have $x=a^2$ and $y=b^2$. This is why $a^2$ is even. You can further deduce that $a$ is even from here if you choose.
A: You are given:$$a^2=2b^2$$This implies that $a^2$ is equal to two times some number.
If you multiply any integer by $2$ then you will get an even number.
We can therefore conclude that $a^2$ must be even.
Next we observe that the only way to get an even integer after squaring is to start off with an even integer.
In other words: even times even = even
Where as: odd times odd = odd
We can therefore conclude that since $a^2$ is even then this implies that $a$ must also be even.
A: Fundamental Theorem of Arithmetic says that every integer $>1$ can be uniquely expressed as a product $p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ (for primes $p_i$ and integers $\alpha_i$).  
Since you know $a^2=2b^2$, you know that $2$ has to be somewhere in the representation of $a^2$, and thus in the representation of $a$ (raising $a$ to any positive integer power $n>1$ only enlarges the $\alpha_i$, not changing the primes $p_i$).  
By definition, a number $b$ is even iff $2$ is somewhere in the unique prime-product representation of $b.$
A: a and $a^{2}$ have the same parity. if $a$ is even, then $a=2k$ for some integer $k$. hence $a^{2}=4k^{2}$, still even!. And if $a$ is odd, a=2k+1, $a^{2} =4k^{2}+4k+1=2(2k^{2}+2k)+1$ so $a^{2}$ is odd! (it also follows from modlar arithmetic)!
A: $a^2=2b^2$ means $a^2$ is divisible by 2. So I think you want to prove if $a^2$ is even, then $a$ is even. Or we can prove the contrapositive. If $a$ is odd, then $a^2$ is odd. $a=2k+1$ implies $a^2=4k^2+4k+1=2(2k^2+2k)+1$ which means $a^2$ is also odd. 
A: If $a$ is odd, so is $a^2$ and so is $2b^2$ !?
