If a^2+b^2=c^2 (a^2 represents 'a' squared), then 2a, 2b and 2c also satisfy the Pythagorean equation, meaning (2a)^2+(2b)^2=(2c)^2. This means that you can have a Pythagorean triangle with all sides even.

Let a , b , c be the three sides of the Pythagorean triangle such that a2 + b2 = c2.

Let two sides are even

If a = 2p and b = 2q , then 4 (p2 + q2) = c2 .

Since LHS is divisible by 2 therefore RHS is also divisible by 2 i.e. c is divisible by 2, i.e. c is also even.

If a = 2p and c = 2q , then b2 = c2 - a2 = 4(q2 - p2) , again by the same argument b must be even.

Thus, if two sides of a Pythagorean triangle are even, then the third side is also even. Hence all the sides are even.

Next, we show that there is at-least one side in the Pythagorean triangle, which is even.

If Possible, let all the sides be odd, then b2 = c2 - a2 = (c+a)(c-a) = even number [ as difference of two odd number is even]. Thus b is even, a contradiction.

Thus, either all the sides of the Pythagorean triangle are even or at-least one side is even.