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axl_oz
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte Updated on: 28 Jul 2025, 08:32
Originally posted by axl_oz on 15 Dec 2009, 15:48.
Last edited by Bunuel on 28 Jul 2025, 08:32, edited 1 time in total.
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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
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D
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 15 Dec 2009, 17:04
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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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
Show more
Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2.

\(2^4+2^2=4^2+2^2\) --> \(|4-2|=2\)

But if we check for \(y=0\), we'll get:

\(2^x+2^0=x^2+0^2\) --> \(2^x+1=x^2\) --> \(2^x=(x-1)(x+1)\) --> \(x=3\)

\(2^3+2^0=9=3^2+0^2\)

\(|x-y|=|3-0|=3\)

4 cannot be the greatest value as when you increase \(x\) so as \(x-y\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\).­
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 08 Feb 2013, 09:27
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axl_oz
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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
Show more

Since we need to maximize 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.
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte Updated on: 11 Oct 2022, 02:24
Originally posted by KarishmaB on 21 Mar 2013, 21:56.
Last edited by KarishmaB on 11 Oct 2022, 02:24, edited 1 time in total.
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Brunel and all,

Is it a rule to apply one value as zero whenever it is given:

1) Both x and y are non-negative integers
2) we need to find the max value of x-y

What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?
Show more

Usually, when you are checking for numbers, you do check for 0. It's often a transition point for patterns. Secondly, the question used the term 'non-negative integers' instead of 'positive integers' - this means 0 would probably have a role to play.
There are no such rules but common sense says that we must not ignore 0.

Also, this question doesn't really test max min concepts. It is a direct application of your understanding of exponential relations.

Now, when we look at the equation, 2^x + 2^y = x^2 + y^2, some things come to mind:
1. It is not very easy to find values that satisfy this equation.
2. But there must be some values which satisfy since we are looking for a value of |x – y|
3. If x = y = 2, the equation is satisfied since all terms become equal and |x – y| = 0 which is the minimum value of |x – y|.

Usually, the left hand side will be greater than the right hand side (as discussed in the post, 2^n will usually be greater than x^2 except in very few cases). So we must focus on those 'very few cases'. Also, we need to make x and y unequal.

We know (from the post) that 2^4 = 4^2 is one solution so we could put x = 4 while keeping y = 2. The equation will be satisfied and |x – y| = 2

Now, we also know that 2^x < x^2 when x = 3. So is there a solution there as well? The difference between 2^3 and 3^2 is of 1 so can we create a difference of 1 between the other two terms? Sure! If y = 0, then 2^0 = 1 but 0^2 = 0.
So another solution is 2^3 + 2^0 = 3^2 + 0^2.
Here, |x – y| = 3 which is the maximum difference.

The reason we can be sure that there are no other values is that as you go ahead of 4 on the number line, 2^n will be greater than n^2 (again, discussed in the post). So both left hand side terms will be greater than the right hand side terms i.e. 2^x > x^2 and 2^y > y^2. So, for no other values can we satisfy this equation.

Answer (D)
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 19 Dec 2009, 07:20
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Great Problem! Since your trying to find the greatest value of X-Y, you just have to assume that Y=0, like Bunel said and then use the "hit and trial" approach like xcusem... Said. The algebratic approach is great too, but I know for me personally it opens up the opportunity for me to make silly mistakes. So I try to not use it unless necessary.

Posted from my mobile device
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 21 Mar 2013, 09:34
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Brunel and all,

Is it a rule to apply one value as zero whenever it is given:

1) Both x and y are non-negative integers
2) we need to find the max value of x-y

What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 14 Oct 2013, 12:10
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the expression |x-y| to reach its maximum we need y to be 0. Hence, we need to find X.

therefore, 2^x+2^y=x^2+^2 --> 2^x+1=x^2 what means that x is odd. Only 3 satisfies this equation: 2^3+1=3^2.
Hence, x must be equal 3
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 06 Dec 2013, 10:47
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axl_oz
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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)

my answer:
|x – y| for x and y 2^x + 2^y = x^2 + y^2
A. x= y x=y=1 2 + 2 = 1 + 1 wrong
B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong
C. 2 x=3, y=1 8 + 2 = 9 + 1 right
D. 3 x=4, y =1 16 + 2 = 16 + 1 wrong
E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong

My answer is C.

Please note that the correct answer is D, not C.
Show more

Hi thank you for your reply.
I explained what i confused. Of course I read previous answers and all chose D.
However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer.
Please help!
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 07 Dec 2013, 05:52
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my answer:
|x – y| for x and y 2^x + 2^y = x^2 + y^2
A. x= y x=y=1 2 + 2 = 1 + 1 wrong
B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong
C. 2 x=3, y=1 8 + 2 = 9 + 1 right
D. 3 x=4, y =1 16 + 2 = 16 + 1 wrong
E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong

My answer is C.

Please note that the correct answer is D, not C.

Hi thank you for your reply.
I explained what i confused. Of course I read previous answers and all chose D.
However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer.
Please help!
Show more

To get the greatest value of |x-y| as 3 consider x=3 and y=0. Notice that these values satisfy \(2^x + 2^y = x^2 + y^2\) --> \(2^3 + 2^0 =9= 3^2 + 0^2\).

Hope it helps.
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 01 Sep 2015, 11:05
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[quote="dharam831"]If 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers, what is the greatest possible value of |x – y|?

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

hi,

It is given that |x-y| should be maximize and both x and y can not be negative (they are non-negative integers). This means to maximize |x-y| y value should be zero.
Now if y=0, then the equation will be :
2^x + 2^0 = x^2 + 0^2
=> 2^x +1 = x^2
=> x^2 - 2^x = 1
This is possible in only one condition when 3^2 - 2^3. and hence x = 3.
So, maximum value of |x-y| is |3-0| = 3 Ans(D).
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 06 Mar 2018, 15:09
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Hi All,

This question is layered with clues to help us solve it, but the work that we'll do is more "brute force" than anything else.

We're given 3 pieces of information:

1: 2^x+2^y=x^2+y^2
2: X and Y are NON-NEGATIVE integers (this means that they're either 0 or positive)
3: The answers are small integers (0 - 4, inclusive)

We're asked for the GREATEST possible value of |X-Y|…..

From the answers, we know that X and Y have to be relatively "close" on the number line. Next, the phrase "non-negative" is interesting - it gets me thinking that one of the values is probably going to be 0.

Now, let's do a few brute force calculations so we can see the results:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16

0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16

According to the given equation, we need to sum two values from the first list and sum the two corresponding values from the second list (and have the results equal one another). There are a couple of ways to do that, but we want the GREATEST 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

Final Answer:
Show Spoiler
D


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Rich
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 21 Aug 2021, 19:57
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using graph
y can only be zero as minimum value for maximum contribution in |x-y|

lets see what could be max value of x ; now set y=0

2^x+ 1= x^2
if x=0 then LHS>RHS
if x=1LHS>RHS
if x=2 LHS>RHS
if x=3
9 = 3^2

Hence X= 3

final value of mode |x-y| = 3-0= 3

hence final answer D

Bunuel VeritasKarishma @chetan4u zhanbo please comment on this approach
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 23 Aug 2021, 23:04
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mSKR
using graph
y can only be zero as minimum value for maximum contribution in |x-y|

lets see what could be max value of x ; now set y=0

2^x+ 1= x^2
if x=0 then LHS>RHS
if x=1LHS>RHS
if x=2 LHS>RHS
if x=3
9 = 3^2

Hence X= 3

final value of mode |x-y| = 3-0= 3

hence final answer D

Bunuel VeritasKarishma @chetan4u zhanbo please comment on this approach
Show more

Yes, familiarity with relation between 2^x and x^2 helps.
We know that the graph of 2^x lies below x^2 for very few values.
There are two points where the two graphs intersect, x = 2 and x = 4.

Between 2 and 4, 2^x < x^2.
So 2^x can be less than x^2 (if 2^x must be equal to x^2 - 1) only between 2 and 4.

Put x = 3 and we get 2^3 = x^2 + 1.

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/01 ... cognition/
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 09 Jul 2023, 11:11
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axl_oz
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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
Show more


My take on this question,
\(2^x + 2^y = x^2 + y^2\)
\(2^x + 2^y = x^2 + y^2 -2xy +2xy \)
\(2^x + 2^y -2xy = (x-y)^2\)

Square root both sides,
\(\sqrt{2^x + 2^y -2xy} = |x-y| (\sqrt{x^2} = |x|)\)

For RHS to be maximum, y= 0 or -ve(not possible)

\(\sqrt{2^x + 2^y -2xy} \)
\(\sqrt{2^x + 2^0}\)
\(\sqrt{2^x + 1}\)

only perfect square possible is when x=3
\(\sqrt{2^3 + 1}\)
\(\sqrt{9}\)

|x-y| = 3

Bunuel Can this be done in this way ?

Option D
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 01 Sep 2024, 19:41
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It is imortant to note that x or y will never be negative integer as it will make the LHS non integer, except when both x and y are -1.
However, (x,y) = (-1,-1) doesnt satisy the equation.

Hence, x and y are positive integer. now We can maximize by making one integer 0.

if y is zero, 2^x +1 = x^2 clearly x = 3 satisfies the equation and hence 3 is max value of mod(x-y) when (x,y)= (3,0)

P.S x = 4 doesnt satisfy the equation. ;)
axl_oz
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

Show Spoiler
This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)­
Show more
­
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 05 Sep 2024, 02:30
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axl_oz
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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2.

\(2^4+2^2=4^2+2^2\) --> \(|4-2|=2\)

But if we check for \(y=0\), we'll get:

\(2^x+2^0=x^2+0^2\) --> \(2^x+1=x^2\) --> \(2^x=(x-1)(x+1)\) --> \(x=3\)

\(2^3+2^0=9=3^2+0^2\)

\(|x-y|=|3-0|=3\)

4 can not be the greatest value as when you increase \(x\) so as \(x-y\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\).
Show more
Hello! How do you get this?
\(2^x=(x-1)(x+1)\) --> \(x=3\)­
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 06 Sep 2024, 04:31
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Hi

There is one step which will clarify your doubt.

\(2^x+1=x^2\)
Quote:
\(2^x=x^2-1 \)
Show more
\(2^x=(x-1)(x+1)\)

I hope it is clear. Feel free to tag me in case of any query.
lnyngayan

Bunuel

axl_oz
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

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2.

\(2^4+2^2=4^2+2^2\) --> \(|4-2|=2\)

But if we check for \(y=0\), we'll get:

\(2^x+2^0=x^2+0^2\) --> \(2^x+1=x^2\) --> \(2^x=(x-1)(x+1)\) --> \(x=3\)

\(2^3+2^0=9=3^2+0^2\)

\(|x-y|=|3-0|=3\)

4 can not be the greatest value as when you increase \(x\) so as \(x-y\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\).
Hello! How do you get this?
\(2^x=(x-1)(x+1)\) --> \(x=3\)­
Show more
­
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GMAT 1: 650 Q56 V55
Schools: Kelley '26
GMAT Focus 1: 615 Q82 V71 DI66
GMAT 1: 650 Q56 V55
Posts: 17
Kudos: 2
If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte 08 Oct 2025, 07:56
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thanks for the explanation
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