I understand this is intuitively obvious. I am wondering if there is a axiom or principle or proof that shows this? It seems you may need something that says you can decompose an integer into it's lowest common factors and if a number is odd then each of it's lowest common factors will never be divisible by two which is the lowest possible even integer.
It seems like there may be something fundamental beside my intuition here in number theory maybe...and that I either forgot or never learned.
Let $n$ be an odd integer. Then, there exists an integer $k$ such that $n = 2k + 1$. Thus, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$. Since there exists an integer $l$ such that $n^2 = 2l + 1$, namely, $l = 2k^2 + 2k$, $n^2$ is odd.
An odd number can be represented as $2k{+}1$ with $k\in \Bbb Z$ (and an even number cannot). So squaring that expression $(2k+1)^2=4k^2+4k+1 = 2(2k^2+2k) + 1$ and clearly $(2k^2+2k)\in \Bbb Z$ so the square is odd.
Equivalently, you could work $\bmod 2$ since any odd number $n\equiv 1 \bmod 2$ and thus $n^2\equiv 1^2\equiv 1 \bmod 2 $ is odd also.
Every number has a unique prime factorization (fundamental theorem of arithmetic). This means any odd number, $a$ is
$a = f_1^{p_1} * f_2^{p_2} * ... * f_N^{p_N}$
where none of the $f_i = 2$.
You can write $a^2$ in terms of the $f_i$ and see that
$a^2 = f_1^{2p_1} * f_2^{2p_2} * ... * f_N^{2p_N}$,
and again no prime factors are 2.
I.E. squaring does not change the spectrum of prime factors, only their powers, so it can't take a number from odd to even.
1) The product of 2 odd numbers is odd.
$a=2k+1; b= = 2l+1.$
$ab= (2k+1)(2l+1) =$
$ 4kl + 2(k+l) +1 =$
$ 2[2kl +(k+l)] +1$, odd.
In your case $a=b$, hence $a^2$ is odd.
2) The product of 2 even numbers is even.
$a=2k; b=2l;$
$ab= 2(2kl)$, even.
3) The product of an even and an odd number is even.
$a=2k; b=2l+1;$
$ab= 2k(2l+1)= 2[k(2l+1)],$ even.
4) How about the product of an odd and an even number? :)
It is possible to use the definition of a prime here - assuming you are allowed to know that $2$ is prime.
So $p$ is a prime if $p|mn$ implies that $p|m$ or $p|n$.
(In some approaches to arithmetic this is a theorem rather than a definition - or rather the theorem is that over the integers every irreducible is prime. Sometimes the properties of an irreducible are used as the definition of integer primes. It is only later that the distinction starts to become important.)
Apply this to $2|n^2$ to conclude that $2|n$ or $2|n$ i.e. $n$ is even.
Observe that $$ m^2=2\binom{m+1}{2}-\binom{m}{1}=2\left(\frac{m(m+1)}{2}\right)-m. $$ When $m$ is odd $m^2$ thus equals an even number minus an odd number which is odd.
Do you know that you can represent an odd integer using the following form: $2k + 1$ where $k=0,1,2,...$. In other words, $2k + 1$ in effect says that an odd integer is nothing more than an even integer plus one. So, what this all means is that if you pick any integer $k$ whatsoever from the set $\{0,1,2,...\}$, plug it in into the formula, as a result you get an odd number.
What if we now square it? Well, we get the following: $(2k + 1)^2=(2k)^2+2\cdot 2k \cdot 1 + 1^2=$ $4k^2+4k+1$. Now, placing everything else except the $1$ into a set of parentheses and factoring out a $2$, we get: $2(2k^2+2k)+1$. What you now observe is the fact that $2(2k^2+2k)+1$ is nothing but an even integer plus one which means that it's odd. This in turn means that the expression $(2k + 1)^2$ (the square root of an odd integer) we started with was an odd number to begin with. And that's how you prove that if an integer is odd, its square is an odd number too.
Hm... I'm not sure whether I'll be able to write this one as I have it in my mind. Let's think of this as the answer for dummies:
a*b = a+a+a+a+a+...+a where we have b amount of a.
a^2 = a * a = a+a+a+a+a+...+a where we have a amount of a.
Let's consider a=(2k+1) - therefore an odd number, where k is of {0,1,2,3,...,inf}
Therefore we can write
a * a == (2k+1) * (2k+1) == 2k(2k+1)+ 1*(2k+1) == 4(k^2)+2k + (2k + 1) == 4(k^2) + 2k + 2k +1
Here we can by now clearly see, that whatever number k is, each has either a 4 or a 2 next to each other making them even because even*odd = even [ 1+1 = 2*1 = 2 ]
Note: I just realized that Peter Szilas wrote it a lot shorter than I did... I didn't actually rip off his answer. It occurred to me at this point of writing, that someone most probably thought of this as it's fairly elementary.
Back to the problem:
4(k^2) + 2k + 2k +1 == even + even + even + odd
Since there is only a single odd number in this sum, the result will be odd.
I'm aware that there is a more sophisticated way of explaining this, but It's been a long time since I used Math and English at the same time.
Edit: Actually this is a sub-problem of if a is odd b is odd then a * b = odd
... and someone else explained this as well as I see. Never mind.
Let $m$ be an odd integer. Suppose $m^2= 2k,$ with $k$ is an integer. Then
$$m\left(\frac{m}2\right)=k,$$ giving a contradiction, since $m$ has no prime factors as $2$.
