Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists. 
Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.

I tried many different numbers and nothing is working but I don't know how to prove that it isn't possible.
 A: $$64 = 17^2-15^2 = 10^2-6^2 = 8^2 - 0^2$$
Unless you are also counting $(-17)^2 -(-15)^2$, $(-17)^2-15^2$, etc. as different ways. 

Don't like subtracting $0^2$? 
$$256 = 65^2-63^2 = 34^2-30^2 = 20^2-12^2$$
A: I assume you mean "difference of squares of positive integers". 
Let $n$ be any positive integer, and $x,y$ positive integers such that $n = x^2 - y^2$. Then $n = (x - y)(x + y)$, so we get a factorization of $n$ into two integers. On the other hand, if we have a factorization $n = ab$, we can get $x$ and $y$ by $x = \frac{a + b}{2}$ and $y = \frac{b - a}{2}$. To get positive integers, we need $b > a$, and also for both $a$ and $b$ to have the same parity. (Both even or both odd). So we get:

The number of ways of writing $n = x^2 - y^2$ where $x$ and $y$ are positive integers is the same as the number of ways of writing $n = ab$, where $a$ and $b$ are positive integers with the same parity and $a < b$.

Now, if $n$ is even, we need both $a$ and $b$ to be even, meaning $n$ is divisible by $4$. Then factorizations $n = ab$ when $a$ and $b$ are both even are the same as factorizations $n/4 = a'b'$. So we get:

If $n$ is divisble by 4, the number of ways of writing $n = x^2 - y^2$ where $x$ and $y$ are positive integers is the same as the number of factorizations $n/4 = a'b'$, where $a' < b'$. (Explicitly, we can take $x = b' + a'$ and $y = b' - a'$)

There are two cases here: If $n/4$ is not a perfect square, the number of factorizations is half the number of factors. Thus we are looking at numbers with exactly $6$ factors. These are numbers of the form $p^5$, where $p$ is a prime, or $pq^2$, where $p$ and $q$ are distinct primes. If $n/4$ is a perfect square, the number of factorizations into distinct integers is half of one less than the number of factors, so we are looking at perfect squares with 7 factors. These are of the form $p^6$, where $p$ is prime. So there are three "families" of solutions:
$$n = 4pq^{2}; \qquad n = 4p^{5}; \qquad n = 4p^{6}$$
The smallest solution is therefore $4 \cdot 3 \cdot 2^{2} = 48$. The factorizations of $3 \cdot 2^{2}$ are $1 \cdot 12$, $2 \cdot 6$, $3 \cdot 4$. This yields the three solutions:
$$48 = 13^{2} - 11^{2} = 8^{2} - 4^{2} = 7^{2} - 1^{2}$$
A: I found $80$ 
$80=9^2-1^2=12^2-8^2=21^2-19^2$
A: All the multiples of 4 except 4 can be represented as a difference of squares of positive integers a and b where a = n/4 + 1 and b = n/4 - 1.
