If 2^x + 2^y = x^2 + y^2, Where x and y are Nonnegative GMAT Problem Solving

Question: 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

“If 2^x + 2^y = x^2 + y^2, Where x and y are Nonnegative GMAT Problem Solving”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “Guide Official Guide Quantitative Review 2022”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:
Approach Solution 1

|x-y| is given. Must be maximised and both x and y must not be negative (they are non-negative integers). This means |x-y|. To maximise the y value, it should be zero.

For y = 0, the equation becomes

2^x + 2^0 = x^2 + 0^2
=> 2^x +1 = x^2
=> x^2 - 2^x = 1

This is only possible under one condition: 3^2 - 2^3. So x = 3.

Therefore, the maximum value of |x-y| is |3-0| = 3

Correct Answer: D

Approach Solution 2

Two values from the first list should be summed, and two corresponding values from the second list should be summed according to the given formula (results should be equal to each other). There are several ways to do this, but you want the largest possible difference between X and Y....

If X = 3 and Y = 0, then we'd have

2^x+2^y=x^2+y^2
8 + 1 = 9 + 0
9 = 9
|3 - 0| = 3

Correct Answer: D

Approach Solution 3

Since we need to maximise the value of |x – y|, we can do that in two ways...1)make y negative, which is not possible as per the question...2)make y= 0..putting y=0 you will get an equation in x and on hit and trial method u will get the value of x as 3, which will satisfy the equation....

putting x=3 and y=0, we will get the value of |x – y| as 3.
Maximise |x-y| by either making y negative or y = 0. y cannot be negative as given, so make y = 0.
Plug 0 in for y to get x^2 - 1 = 2^x
x^2 -1^2=2^x
If two integers have a median value, then they have a difference of squares.
We already know that they have a difference of squares, so we need to find the median value.

(x+1)(x-1)=2^x
x can only be odd numbers for there to be a median integer.
Plug 3 and you get 4*2 = 2^3.
That works, so x=3.

Correct Answer: D

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