Given $x$ and $y$ are multiples of $2$ satisfying $$x^2 - y^2 = 27234702932$$ Find the number of solutions to $x$ and $y$.
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As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes
If $X^2-Y^2=p\cdot q$ where $p,q$ are primes,
the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$ and $\pm p,\pm q $
For example, if $X+Y=1,X-Y=pq$
The number of positive factors of $p\cdot q$ is $(1+1)\cdot(1+1)=4$
So, there should be $4\cdot2 =8$ solutions in integers.
First factor $27234702932=2^2\times181\times37616993$. Since both are even, so is $x+y$ and $x-y$. Each solution is uniquely determined by $x+y$ and $x-y$.