Find a pythagorean triple where each integer is odd?

Answer 1

No such triple exists!
If x,y,z is a primitive Pythagorean triple recall that
x = u²-v²
y = 2uv
z = u²+v²,
so at least one of x,y,z is even.
If the triple is not primitive, we get the same
result, with each of x,y,z multiplied by the
same factor t, so the same result holds.

Second proof: If x and y are both odd
then x² = y² = 1(mod 4). Since x²+y² = z²,
That says z² = 2(mod 4), which is impossible.

Answer 2

There is no such tripple.

a^2 + b^2 = c^2

(odd number) ^2 = another odd number (3 X 3 = 9, 7 X 7 = 49)
add 2 odd numbers you get an even number.

even number can not be square of odd number.

Answer 3

I could not find one. Are you sure there is even one?

Answer 4

i admit i couldn't find any one of them must be even sorry hope someone out there finds