How many?

Relevant wiki: Parity of Integers

Solution inspired by Prithwish Roy

$\begin{aligned} x^2 - y^2 & = 353702 \\ (x-y)(x+y) & = 353702 & \text{for }x > y \end{aligned}$

There are only three cases to consider.

• If either $x$ or $y$ is odd and the other even , then both $x-y$ and $x+y$ are odd and the LHS is odd, but the RHS is even, therefore, there is no solution .
• If both $x$ and $y$ are odd , then both $x-y$ and $x+y$ are even and the LHS is divisible by 4, but the RHS is indivisible by 4, therefore, there is no solution .
• If both $x$ and $y$ are even , then both $x-y$ and $x+y$ are even and the LHS is divisible by 4, but the RHS is indivisible by 4, therefore, there is no solution .

Therefore, there is $\boxed{0}$ integral solution.

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My earlier solution

$\begin{aligned} x^2 - y^2 & = 353702 \\ (x-y)(x+y) & = \color{#3D99F6}mn & \small \color{#3D99F6} \text{where }m, n, m<n \text{ are factors of }353702. \end{aligned}$

$\implies \begin{cases} x - y = m \\ x + y = n \end{cases} \implies \begin{cases} x = \dfrac {m+n}2 \\ y = \dfrac {n-m}2 \end{cases}$

Since in prime factors $353702 = 2 \times 17 \times 101 \times 103$ , either $m$ or $n$ is even and the other is odd. Therefore, both $m+n$ and $n-m$ are odd and $x$ and $y$ are never integers, and there is $\boxed{0}$ integral solution.

Well the above situation is only possible when both factors of $x^{2}-y^{2}$ that are $x+y$ and $x-y$ , are even. Thus the number 353702 must be divisible by 4. But it isn't. So there aren't any real solutions.