<s>[INST] Find all even numbers that can be represented as a difference of squares in only two ways I am currently working on this proof. I am looking to find (with proof) all even numbers that can be represented as a difference of squares in only two ways.
My thoughts thus far.
I examined the first 40 natural numbers and found that 16, 24, 32, 40 have exactly two representations as a difference of squares. My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations. 
$8k+16=(k+a)^2 - (k+b)^2$ $\rightarrow$ $8k+16=(k+4)^2 - k^2$ which is the first representation.
My idea for finding the second representation is to express $8k+16$ as
$8k+16=(ak+b)^2 - (ck+d)^2$
Expanding and attempting to solve I get:
$$\left.\begin{matrix}
 a^2 - c^2 = 0\\ 
2ab - 2cd= 8\\
b^2 - d^2= 16
 \end{matrix}\right\}$$
This is where my problem arises, $b^2-c^2=(b-c)(b+c)=16=4^2$, I cannot seem to move on from here.
If anyone can offer a hint that would help with my understanding of the problem and the attempt at the proof, that would be extremely helpful. Thank you.
 [/INST] 
My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.

Counterexample: 
\begin{align}
120 &= 31^2-29^2 \\
&= 17^2 - 13^2 \\
&= 13^2-7^2 \\
&= 11^2-1^2
\end{align}

Note that while I just picked a number divisible by $8$ with plenty of factors, the claim actually falls at $48$:
\begin{align}
48 &= 13^2-11^2 \tag{from (12,1)}\\
&= 8^2 - 4^2 \tag{from (6,2)}\\
&= 7^2-1^2 \tag{from (4,3)}\\
\end{align}
because $48/4=12$ has three pairs of factors.
</s>