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are there any positive integers k,a, b such that this equation is satisfied:
k^2-1=a^2+b^2
k^2-1=a^2+b^2
Aditya89 said:Here's the solution.
We have:
k^2-a^2=b^2-1
(k+a)(k-a)=b^2+1
Then choose b such that b is even.
This implies b^2+1 is odd. If b^2+1 is a prime, then put k+a=b^2+1 and k-a=1. You will get k&a. If b^2+1 is not a prime, then choose k+a and k-a as its two odd factors. Solving, you get k & a.
As there is solution for all b>0 and b even, there are infinitely many solutions.