Lest not taking up a lot of time, I would plug in numbers as that would seem fastest ... sadly I don't have a more mathematical approach within the 2 min time frame.

Remember 0 is a +ve Integer.
So I would start with D. Diff 3. Let's say x=3, y =0. Then 2^x+2^y=8+1=9 ; x^2+y^2=3^2+0=9. Others don't satisfy.

IMO D.

If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x - y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

|x-y| = the DISTANCE between x and y.

To MAXIMIZE this distance, try to MAXIMIZE x and MINIMIZE y.
Note the word in red, which implies that y can be equal to 0.
If y=0, we get:
2^x + 2� = x² - 0²
2^x + 1 = x²
x² - 2^x = 1.

The answer choices imply that the distance between x and y cannot be greater than 4.
If y=0, then x must be equal to one of the following values: 0, 1, 2, 3, 4.
Only x=3 satisfies the equation x² - 2^x = 1:
3² - 2³ = 1.

Thus, x=3 and y=0 satisfy the equation 2^x + 2^y = x^2 + y^2.
In this case, |x-y| = |3-0| = 3.

Given that 2^x + 2^y = x^2 + y^2, if the value of y INCREASES -- if y is equal to an integer GREATER THAN 0 -- then the value of x will have to DECREASE.
The result is that x and y will be brought closer together, DECREASING the distance between them.
Thus, the maximum possible distance between x and y is 3.

The correct answer is D.

ani781 wrote:Lest not taking up a lot of time, I would plug in numbers as that would seem fastest ... sadly I don't have a more mathematical approach within the 2 min time frame.

Remember 0 is a +ve Integer.
So I would start with D. Diff 3. Let's say x=3, y =0. Then 2^x+2^y=8+1=9 ; x^2+y^2=3^2+0=9. Others don't satisfy.

IMO D.

Ani your solution is nice. But, I would like to correct you on one statement in your solution.

"0" is NOT positive.. here, we selected "0" because question says that numbers are "non-negative"

This is NOT an easy one! Here's an attempt at a definitive explanation, but it's time-consuming and technical (the number plugging strategies recommended above would at least give you a 50/50 shot between D and E, which is probably better on test day).

Rewrite the equation as 2ˣ - x² = y² - 2ʸ

There are three possibilities for the right hand side of the equation (the easier one to work with):

y² - 2ʸ = 0
y² - 2ʸ < 0
y² - 2ʸ > 0

Let's deal with each one.

Suppose y² - 2ʸ = 0. In this case, y = 2 or y = 4. 2ˣ - x² = 0 also implies x = 2 or x = 4, so in this case our maximum value of |x - y| is |4 - 2| or |2 - 4|, either of which = 2.

Suppose y² - 2ʸ < 0. Then we also have 2ˣ - x² < 0, which only happens if x = 3 and which gives 2ˣ - x² = -1. That means y² - 2ʸ = -1, which happens only if y = 0 or y = 1. So we could have |x - y| = |3 - 0| = 3.

Suppose y² - 2ʸ > 0. This happens if y = 3, and gives y² - 2ʸ = 1, which implies that 2ˣ - x² = 1. This happens only if x = 0 or x = 1, so again we could have |x - y| = |0 - 3| = 3.

We've exhausted all our options, so the maximum is 3.