Is it true that a number of the form $2p$, where $p$ is prime, cannot be written as $a^2-b^2$ for some $a,b\in\mathbb Z^+$? While studying number theory, I thought of a conjecture that I don't know whether it is true or false. 
Conjecture: Let $c$ be a composite number with only two distinct prime factors, $i$ and $j$. If the value of $i + j$ is odd, then $c$ cannot be expressed in the form $c = a^2 - b^2$, where $a,b\in\mathbb Z^+$.
Example: $6$ cannot be expressed as the difference of two squares as the sum of the two distinct prime factors of $6$, $2$ and $3$ is odd. 
Solving two different systems of linear equations: $a + b=3$, $a - b=2$ and $a + b=2$, $a - b=3$ gives either negative or non-integer values of $a$ and $b$, thus it satisfies the conjecture.
If the conjecture is true, how do I prove it? If not, what contradicts it? 
 A: You state $c$ is a composite number which is a product of just two distinct prime factors $i$ and $j$. Since $i + j$ is odd, this means that either $i$ or $j$ is even, i.e., $2$ as it's a prime, and the other one is an odd prime. In that case, WLOG, let $i = 2$, so $c = 2j$, where $j$ is the odd prime. Also, this means $c$ is even.
However, since $c = a^2 - b^2$, then $a^2 - b^2$ must be even, but this means it has a factor of $4$ (since squares have a remainder of $0$ or $1$ when divided by $4$, then both $a^2$ and $b^2$ must have the same remainder, so their difference is divisible by $4$), i.e., it has at least $2$ factors of $2$. This doesn't match the requirement that $c$ has just one factor of $2$ and, thus, is not possible. This shows your conjecture is true.
A: Since $i+j$ is odd, one of $i,j$ is even and the other is odd; since $i,j$ are prime, one of them must be the only even prime $2$. Say $i=2$, so $c=2j$.
Now $c=a^2-b^2=(a+b)(a-b)$. Since $a,b$ are integral, it is easy to show that $a+b$ and $a-b$ must be both even or both odd, so $c$ must also be decomposable as a product of two even or two odd numbers. But this is impossible – the only such decompositions are $c=1×2j=2×j$, both of which have one even and one odd factor. Hence $c$ is not expressible as $a^2-b^2$.
A: This conjecture is true. Indeed, more generally, if $c=2m$ where $m$ is any odd number, then $c$ cannot be written as $c=a^2-b^2$. (This is a generalization because the sum of two primes is odd if and only if one of those primes equals $2$.)
Suppose $c=a^2-b^2$, so that $c=(a+b)(a-b)$. Note that $a+b$ and $a-b$ have the same parity (their difference is $2b$). If both $a+b$ and $a-b$ are odd then their product $c$ is also odd, a contradiction to $c=2m$. And if both $a+b$ and $a-b$ are even then their product $c$ is a multiple of $4$, also contradicting $c=2m$ with $m$ odd.
A: If the number is composite with only two distinct prime factors and the sum of prime factors is odd, implying one factor is 2. But you said that only two distinct prime factors, not -only two factors. So take 24, it has two distinct prime factors- 2,3 which are distinct and 24=4 * 6=(5-1)(5+1). So (a, b) =(5, 1). So your conjecture fails.
