$20172017^2 + b^2 = c^2$

How many positive integer solutions $(b,c)$ exist for this equation?

The answer is 13.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Note that $20172017^2=73^2\times 137^2\times 2017^2$ , and hence $20172017^2$ has $(2+1)(2+1)(2+1)=27$ positive factors. Also, $(c-b)(c+b)=20172017^2,$ and so we are looking for two positive factors $(c-b)$ and $(c+b)$ whose product is $20172017^2$ . There are $27$ ways to assign these factors, but $13$ will give $b$ negative, and one will give $b=0$ (when $(c-b)=(c+b)=20172017$ ). This leaves $13$ ways to assign the factors so that $b$ and $c$ are positive, meaning there are $\boxed{13}$ positive, integral solutions for $(b,c)$ .