Pythagoras in trouble

None of these are really right. If $x,y,z$ have no common factor, then exactly one of $x$ and $y$ is even (which can be seen by a mod-4 analysis). But we can multiply them all by $2$ and get them to be even. You should probably fix the problem to specify that $(x,y,z)$ is a primitive Pythagorean triple.

Here's the solution.

Why can't they both be even?

If x and y are both even, then x^2+y^2=z^2 must be an even number too. And since z^2 is a perfect square, it has to be a multiple of 4 in order to be even, making it of the form, z^2=4.(m^2), where m is some positive integer. So, z=2m, is an even number, which makes x, y and z all even and hence they can't be 'co-prime'.

Why can't they both be odd?

If x and y are odd, x^2 and y^2 are also odd and since the sum of two odd numbers is always even, z^2=x^2+y^2 must be an even number and by our former logic, it has to be a multiple of '4'. Now each odd number can be written in the form '1+an even number', let x=p+1, y=q+1, which makes z^2=x^2+y^2= p^2+q^2+2(p+q)+2, if you observe the sum carefully, you'll notice that the first three terms are divisible by four, leaving 2 as the sole term indivisible by '4'. So if we divide 'z^2' by 4 , we'll get a remainder of '2' which contradicts our original authentic deduction that z^2 is a multiple of '4'.

Which leaves us with only one choice, that only one of x and y can be even.

It's simple! Only one of x and y can be even because if they were both even, they would share a common factor of 2 and would therefore not be coprime which is one of the conditions given in the question.

So statement 4 is the correct answer.

Dude thats not right.

Note that there exists a pythagorean triplet such that 6^2+8^2=10^2