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Now, since both \(x\) and \(y\) are integers (and \(x+y \gt 0\)) then \(x-y=2\) and \(x+y=2\) so, \(x=2\) and \(y=0\). Therefore, \((x+y)^{xy}=2^0=1=odd\), so the answer to the question is No. Sufficient. (Note that \(x-y=1\) and \(x+y=4\) is not a valid scenario (solution), since in this case we would have that \(x=2.5\) and \(y=1.5\), which is not possible as given that both unknowns must be integers)

(2) \(2^x+3^y=\sqrt[(x+y)]{25}\). Obviously \(\sqrt[(x+y)]{25}\) must be an integer (since \(2^x+3^y=integer\)) and as \(x+y=integer\) then the only solution is \(\sqrt[(x+y)]{25}=\sqrt[2]{25}=5\) giving \(x+y=2\). So, \(2^x+3^y=5\). From that, two scenarios are possible:

A. \(x=2\) and \(y=0\) (notice that \(x+y=2\) holds true): \(2^x+3^y=2^2+3^0=5\), and in this case: \((x+y)^{xy}=2^0=1=odd\);

B. \(x=1\) and \(y=1\) (notice that \(x+y=2\) holds true): \(2^x+3^y=2^1+3^1=5\), and in this case: \((x+y)^{xy}=2^1=2=even\).

Hi Bunuel, Can we conclude from the below mentioned eqn, (x-y)(x+y)=4 that either (x-y) or (x+y) has to be even? & if (x-y) is even the (x+y) is bound to be even.

Hence (x+y)^xy will be even. But if any of x,y is 0, the result is odd. Kindly help.

Hi Bunuel, Can we conclude from the below mentioned eqn, (x-y)(x+y)=4 that either (x-y) or (x+y) has to be even? & if (x-y) is even the (x+y) is bound to be even.

Hence (x+y)^xy will be even. But if any of x,y is 0, the result is odd. Kindly help.

Yes, since both x and y are integers then for (x - y)(x + y) = even to be true at least one of the multiples must be even. Next, it turns out that if x + y is even, then x - y is even too and vise-versa. After this I lost you. What is your question?
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I mean to say is Since x+y is even, (x+y)^xy will be even, but if any of x,y is 0, the result is 1. So not sufficient???

Thanks!!

Yes but we can deduce much more from (x - y)(x + y) = 4, then just that both x + y and x - y are even. The only case for (x - y)(x + y) = 4 to be true is when x = 2 and y = 0 (remember that we are given that both are non-negative integers), thus (x + y)^(xy) = 1 = odd.
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