Solve k^2-1=a^2+b^2: Find Positive Ints

[AM-GM]

are there any positive integers k,a, b such that this equation is satisfied:

k^2-1=a^2+b^2

 

[0rthodontist]

Yes, for example k = 9, a = 4, b = 8.

 

[AM-GM]

thanks, how do i go about proving that infinite solutions exist?

 

[Aditya89]

The solution

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.

 

[Aditya89]

You can see that Orthodontist's solution is also one of these.

 

[MathematicalPhysicist]

the case with 1 is a special case of
k^2=a^2+b^2+c^2
a good question that might arose from this is what number is greater: the number of pythogrean triplets or the above qudroplets?

 

[matt grime]

neither, they are both the same (countably infinite); it is not a difficulct question, and has indeed already been answered in this thread where it is asserted that there are infinitely many solutions to

k^2-1=a^2+b^2

 

[murshid_islam]

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.

why should $b$ be even?