Is (X^2+Y^2) a square of an integer?
1). x is even, y is odd
2). x and y are odd
Numbers.
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- neelgandham
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lenagmat wrote:Is (X^2+Y^2) a square of an integer?
1). x is even, y is odd
2). x and y are odd
1) 3^2 + 4^2 = 5^2
15^2 + 16^2 != integer ^2
So, you cannot answer the question, with this data !
2) x and y are odd
Let us write x and y in the form of 2a+1, 2b+1 were a,b are integers. When you square and add these numbers, we get the below.
4(a^2+b^2+a+b) + 2 = 2*(2(a^2+b^2+a+b)+1)
Let c = (a^2+b^2+a+b)
Then we get the final expression = 2* (2*c +1) = Odd * 2 , which can never be a square of a number!
Hence option B
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- Anurag@Gurome
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(1) x is even, y is odd.lenagmat wrote:Is (X^2+Y^2) a square of an integer?
1). x is even, y is odd
2). x and y are odd
If x = -2 and y = 5, then x² + y² = 4 + 25 = 29, which is not the square of an integer.
If x = 4 and y = 3, then x² + y² = 16 + 9 = 25, which is the square of an integer.
We do not get a definite answer; NOT sufficient.
(2) x and y are odd.
If x = 1 and y = 3, then x² + y² = 1 + 9 = 10, which is not the square of an integer.
If x = 3 and y = 5, then x² + y² = 9 + 25 = 34, which is not the square of an integer.
So, x² + y² is not the square of an integer; SUFFICIENT.
The correct answer is B.
Anurag Mairal, Ph.D., MBA
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