That x^2 + y^2 is even, given x and y are integers, implies either x is even and y is odd or x is odd and y is even. We can't be certain which of the two integers is odd and which is even. But what we can be certain of is that when we multiply x and y, it will be even.

So only option E must be an even number considering xy results in an even number irrespective of which of the two integers is even.

The right answer is therefore E.

x^2 + y^2 is odd only when one is definitely even and the other one definitely odd

(A) x --> Can be even or Odd --> NO

(B) y --> Can be even or Odd --> NO

(C) x+y --> Even + Odd or Odd + Even = Odd always --> NO

(D) xy+y --> x(y + 1)
--> Case 1: If x = odd, y = even ; x(y + 1) = odd*odd = Odd
--> Case 2: If x = even, y = odd ; x(y + 1) = even*even = Even
--> Two outcomes are possible --> NO

(E) xy = odd*even or even*odd = Even always --> YES

IMO Option E

If x and y are integers and x2+y2 is odd, which of the following must be even?

If x2 + y2 is odd, this can be inferred that one of x2 and y2 is odd, and another is even.
So, we'll get either x is odd, y is even OR x is even, y is odd.

Now, let's see each answer choice on which choice yield EVEN

(A) x - can be both even/odd
(B) y - can be both even/odd
(C) x+y - definitely odd: since if x is even, y is odd and x+y = odd OR if y is even, x is odd, x+y = odd
(D) xy+y can be both even/odd, if x is even, y is odd; xy+y = odd OR if y is even, x is odd; xy+y = even
(E) xy - one of x and y needs to be even, which even * any number = even - CORRECT answer choice

\(x^2+y^2=odd…e+o=odd\)

(A) x: x=(o,e)
(B) y: y=(o,e)
(C) x+y: o+e=o
(D) xy+y: y(x+1)=o(e+1)=o(o)=o
(E) xy: oe=e

Answer (E)

for the given condition to to be valid of x2+y2 is odd
either of them has to be even and odd
so IMO E ; xy will be even

If x and y are integers and x2+y2 is odd, which of the following must be even?

(A) x
(B) y
(C) x+y
(D) xy+y
(E) xy

Raising an integer by any power will not change its odd/even property.
Also note,
Odd + Odd = Even
Even + Even = Even
Even + Odd = Odd
So in order for x^2 + y^2 to be odd, one of the variables has to even and one has to be odd.
Using numbers we can start by x = 2 and y = 3 or x = 3 and y = 2.

A.) x = 2, yes if x = 2 - no if x = 3 (incorrect as it could or could not be even)
B.) same as above
C.) x + y, 2 + 3 = 5 which is odd (incorrect)
D.) xy + y - if x = 2/y = 3, 2(3) + 3 = 6 + 3 = 9 (ODD), if x = 3/y = 2, 3(2) + 2 = 6 + 2 = 8, (EVEN) (could be either so incorrect)
E.) xy - 2(3) = 6 (EVEN)

Note,
Odd * Odd = Odd
Even * Odd = Even
Even * Even = Even

x^2+y^2= Odd, is possible in two case given below:
1. x is odd and y is even
2. x is even and y is odd.

A. Wrong. x can be even or odd.
B. Wrong. y can be even or odd.
C. Wrong. In both the cases, x+y is odd.
D. Wrong. xy+y=Even*odd+Odd=Even+odd=Odd or Odd*even+even=Even.
E. Correct.

X^2 + y^2 = odd
This implies => odd + even = odd.

Means one if them must be ODD and one of them must be EVEN. Since (odd)^2 = odd and (even)^2 = even.

Implies xy = even!

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