Thread: Pythagorean triplet

Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.

Hello, MATNTRNG!

I have a very primitive solution . . .

Find the only Pythagorean triplet, , for which

All Pythagorean triples are generated by: .


We have: .

. . which simplifies to: .


There are only seven solutions.

. .


And only the last triplet has a sum of 1000.

Originally Posted by MATNTRNG

Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.

So

Therefore we want to solve



Complete the square:

Let to get

Factor:

Now we just have finitely many cases to solve. For example solve

I'll let you solve all of these out, but you should only get one solution with , namely

Now plug these into the formulas for to get

The proof of a=m^2-n^2,b=2mn,c=m^2+n^2 assumed that (a,b)=1. Although it's easily seen that the formula still applies if (a,b) is a perfect square, but there's a potential that you'll miss some solution (although here it doesn't happen) if you merely consider a=m^2-n^2,b=2mn