$x^2 + 4xy = 10^{100}$ how many integer solutions there?
Can't find the answer, because there is too big number. Help me please
I know, how to solve it if I have 100 instead of $10^{100}$: 100 is $2^2$ and $5^2$, so I will try to solve like x = 2, x + 4y = $2*5^2$, then x = $2^2$, x + 4y = $5^2$ etc. But because 10^100 is too big, it will take a lot of time to iterate, so here is my problem
I tried to find it like sum of arithmetic progression (because if we have $10^{11}$ instead of $10^{100}$ there will be 192 solutions, $10^{12}$ => 234 solutions and $10^{13}$ => 280.) So $a_1 = -2, d = 4$, but number of solutions is too big and it can't be true.
I think there are the way to solve it easily
You've got the right idea. $x(x+4y)=10^{100}$, so $x$ must be a factor of $10^{100}$. The factors of $10^{100}$ are all of the form $2^a5^b$, where $0\leq a,b\leq100$. That gives $101^2$ factors, but not every one of them will work as $x$. We also need to be able to solve $2^a5^b(2^a5^b+4y)=10^{100}$ where $y$ is an integer (not necessarily positive.)
How does this restrict the values of $a$ and $b$?