Find number of ordered pairs $(a, b)$ satisfying $a^2+b^2=2^3\cdot3^4\cdot5\cdot17^2$ Find number of ordered pairs $(a, b)$ satisfying $a^2+b^2=2^3\cdot3^4\cdot5\cdot17^2$. By rearranging the terms, I found a pair (918, 306). But I wonder if there is a systematic way to solve for the number of pairs? Any hint will be appreciated!


Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica code:
In[1]:=Length[FullSimplify[
  Solve[{a^2 + b^2 == 2^3*3^4*5*17^2, b > a}, {a, b}, Integers]]]

Running the code gives:
Out[1]=12

Looking for the solutions, we can see:
In[2]:=FullSimplify[
 Solve[{a^2 + b^2 == 2^3*3^4*5*17^2, b > a}, {a, b}, Integers]]

Out[2]={{a -> -954, b -> -162}, {a -> -954, b -> 162}, {a -> -918, 
  b -> -306}, {a -> -918, b -> 306}, {a -> -702, 
  b -> -666}, {a -> -702, b -> 666}, {a -> -666, 
  b -> 702}, {a -> -306, b -> 918}, {a -> -162, b -> 954}, {a -> 162, 
  b -> 954}, {a -> 306, b -> 918}, {a -> 666, b -> 702}}

So, we can see that when we have $(\text{a},\text{b})$ where $\text{b}>\text{a}$ there are $12$ solutions to that problem.
