A much easier solution:
For example, (x,y,z) = (3,4,5) works. We can multiply x,y,z by the same integer constant yielding (x,y,z) = (6,8,10), (9,12,15), etc. Hence there are infinitely many solutions.
Hello all,
I'm practicing proofs and I'm stuck. Here it is:
Prove that there are infinitely many solutions in positive integers x, y, and z to the equation x^2 + y^2 = z^2. Evidently I'm supposed to start by setting x, y, and z like this:
x = m^2 - n^2
y = 2mn
z = m^2 + n^2
So then we have:
(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2
m^4 + n ^4 - 2(mn)^2 + 4(mn)^2 = m ^4 + n^4 +2(mn)^2
m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2 . . . Both sides of the equation are equal
2(mn)^2 = 2(mn)^2 . . . Subtract m^4 and n^4 from both sides.
2 = 2 . . . Divide both sides by (mn)^2
Since 2 always equals 2 there are infinite solutions to the equation m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2. Because x^2 + y^2 = z^2 equals m ^4 + n^4 +2(mn)^2 = m ^4 + n^4 +2(mn)^2, x^2 + y^2 = z^2 also has infinite solutions.
Is this a valid proof or am I missing something?
Ah okay. Your solution is correct, but too long. You can just stop when you know both sides of the equation are equal, i.e. . This equality implies that the above holds regardless of your choice of m and n, as long as m > n (so that x is positive).