Is even hypotenuse possible?

Question

Give an example for the equation
a^(2) + b^(2) = c^(2)
where "a" & "b" are both odd natural numbers & thus "c" is an even natural number.

Answer 1

It is not possible, because
If a and b are both odd, and c even
a=2m+1, b=2n+1, c=2k
Substituting in the Pythagorean equation above
(2m+1)^2+(2n+1)^2=(2k)^2
4(m^2+n^2+m+n)+2=4k^2
Right side is divisible by 4, but left side leaves remainder 2 when divided by 4.

Answer 2

It is not possible. Let a = 2m + 1 and b = 2n + 1

a^2 + b^2 = 4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4(m^2 + n^2 + m + n) + 2

So it is congruent to 2 mod 4. Now look at any even number c = 2q.
c^2 = (2q)^2 = 4q^2 is congruent to 0 mod 4, so there is no solution.

Answer 3

never heard of one