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First off, an even number squared will give an even number(2^2 = 4, 4^2 = 16 etc). An odd number squared will give an odd number (3^2 = 9, 5^2 = 25 etc). Also, for the multiplication between two numbers to come out even, at least one of the numbers must be even.
We know all primes except for 2 are odd. Since x and y are positive, the only way for x + 5 to be prime will be if x is even (i.e. if x = 1 then x + 5 = 6 which is not prime. But if x = 2 then x + 5 = 7 which is prime). Therefore, x^2 will also be even and x^2y^2 will also be even regardless of what y is. statement 1 is sufficient.
Using a similar approach for condition 2, we can set y = 1 (y + 1 = 2 which is prime) or y = 2 (y + 2 = 3 which is also prime). So we see with this condition, y can be both even or odd. So statement 2 alone is not sufficient.
In order \(x^2y^2\) to be even at least one of the unknowns must be even.
(1) \(x+5=prime\) --> as \(x\) is a positive integer then this prime can not be the only even prime 2 (in this case \(x+5=2\) --> \(x=-3=negative\)), so \(x+5=prime=odd\) --> \(x=odd-5=odd-odd=even\). Sufficient.
(2) y + 1 is a prime number --> \(y\) could be 1, so odd, and we won't be sure whether \(x^2y^2=even\) or \(y\) could be even (for example 2) and then \(x^2y^2=even\). Not sufficient.
Re: If x and y are positive integers, is x^2*y^2 even ? (1) x + [#permalink]
12 Apr 2015, 21:20
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