Proof of observations on natural numbers being expressed as differences of squares. Inspired by this Hagon Von Eitzen's answer( https://math.stackexchange.com/a/1591028/789547) I
started investigating how I could express natural numbers as differences of squares.
Using the method explained in the solution,
I investigated natural numbers up to $60$ manually.
I observed the following:

*

*There is no way to express numbers of the form $4k+2$ as a difference of two squares.

2.There is one and only one way to express prime numbers
(except $2$) as a difference of two squares.


*The smallest number that can be expressed as a difference of two squares in two different ways is $15$ and the smallest number that can be expressed as a difference of two squares in three different ways is $45$.

As,
$15= 8^2-7^2=4^2-1^2$.
$45=23^2-22^2=9^2-6^2=7^2-2^2$.
I could prove
point $2$ easily but could not prove point $1$.
Also I am interested in knowing how I could find the smallest natural number that could be expressed as a difference of two squares in four different ways without manual calculation.
 A: Suppose a natural number $n$ is the difference of two squares $x^2$ and $y^2$. Then
$$n=x^2-y^2=(x+y)(x-y),$$
so for $a=x+y$ and $b=x-y$ we have $n=ab$, where $a$ and $b$ are natural numbers. Conversely, if $n=ab$ for natural numbers $a$ and $b$ then for $x=\tfrac{a+b}{2}$ and $y=\tfrac{a-b}{2}$ we have
$$x^2-y^2=\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2=ab=n.$$
Of course $x$ and $y$ are not necessarily integers; they are integers if and only if $a+b$ and $a-b$ are both even. Equivalently, if and only if $a$ and $b$ are both odd, or both even.
So the number of ways we can express a natural number $n$ as a difference of two squares is the same as the number of ways we can factor $n$ as a product of two integers, that are both odd or both even. In particular, this shows that if a number is a difference of two squares, then it is either odd or divisible by $4$, so it cannot be of the form $4k+2$.
Can you use this characterization to find the least natural number that is a difference of two squares in four ways?
A: Point 1 Proof:
$$4k+2=2(2k+1)\ =\ a^2-b^2=(a-b)(a+b)$$
Obviouslu LHS is divisible by 2 so $2\mid a-b$ or $2\mid a+b$. It means that $a$ and $b$ are either both even or both odd. So RHS has two factors divisible by 2 and therefore must be divisible by 4. But the LHS is divisible by 2 only (the other factor of LHS is odd). Contradiction.
Alternative proof: It's easy to check that all squares are either equal to 0 or 1 modulo 4. Their differencies can be 0, 1 or 3 but not 2 (modulo 4)
