It is one equation in 2 variables - so there are an infinite number of solutions. Graphically it is a circle with radius sqrt(1225)=35. Are you looking for only integer solutions?
So for a^2 + b^2 = c^2, I knew that (3,4,5) is a Pythagorean Triplet (PT)
So multiplying all of them by 7 will also be a PT. This gives the solution (21,28,35)
This relies on c = 5 being a solution to a^2+b^2=25 where a=3,b=4
There are no PTs having c = 7. So a^2 + b^2 <> 49 (easy to verify by exhaustion as there aren't that many even/odd pairs to check). If there was we could multiply them all by 5 and have a solution to your equation.
So I am going to say that (21,28,35) is the only one: a= 21,b=28,c=35
You can 1) interchange a and b and 2) juggle negative signs using all 3 to get a total of 2(1+3+3+1) = 16 solutions in the integers.
Maybe you can look at squares modulo 8 - I didn't.
I think (even)^2 + (odd)^2 is congruent 1 or 5 (mod 8) but as I said I did not pursue this.
It is one equation in 2 variables - so there are an infinite number of solutions. Graphically it is a circle with radius sqrt(1225)=35. Are you looking for only integer solutions?
@Ian Fowler yes, if you please.
The only think I could come up with is:
1225 = 35^2 = (7.5)^2
So for a^2 + b^2 = c^2, I knew that (3,4,5) is a Pythagorean Triplet (PT)
So multiplying all of them by 7 will also be a PT. This gives the solution (21,28,35)
This relies on c = 5 being a solution to a^2+b^2=25 where a=3,b=4
There are no PTs having c = 7. So a^2 + b^2 <> 49 (easy to verify by exhaustion as there aren't that many even/odd pairs to check). If there was we could multiply them all by 5 and have a solution to your equation.
So I am going to say that (21,28,35) is the only one: a= 21,b=28,c=35
You can 1) interchange a and b and 2) juggle negative signs using all 3 to get a total of 2(1+3+3+1) = 16 solutions in the integers.
Maybe you can look at squares modulo 8 - I didn't.
I think (even)^2 + (odd)^2 is congruent 1 or 5 (mod 8) but as I said I did not pursue this.