Bunuel wrote:

Official Solution:

(1) \(4y^2+3x^2=x^4+y^4\)

Rearrange: \(3x^2-x^4=y^4-4y^2\);

\(x^2(3-x^2)=y^2(y^2-4)\).

Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even, then the product is naturally even. So, \(y^2(y^2-4y)\) is also even, but in order for it to be even \(y\) must be even, since if \(y\) is odd, then \(y^2(y^2-4y)=odd*(odd-even)=odd*odd=odd\). Sufficient.

(2) \(y=4-x^2\)

If \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.

Answer: A

I did it without regrouping.

1) 4y^2 - is alway even; 3x^2 could be even or odd, depending on X;

x^4 will be even or odd, depending on X, but it will be the same as above; y^4 is unknown?

e + e = e => e = e + something, then this something is even

e + o = o => o = o + something, then this something is even

2) Not sufficient