If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte

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)

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.

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I have a question for Bunuel: we know that |x−y| = ((x-y)*2)*1/2 = (x*2+y*2-2xy)*1/2

so substituting the value of x*2 +y*2 as 2*x + 2*y in the above equation

I got |x−y| = (2*x + 2*y -2xy)*1/2

then, substituting values for x (taking y=0)...the max value for x can be anything more than 0, coz if you take (x=4) then u'l end up with |x−y| = 4

please can you help me with this problem, where did I go wrong???

I don't understand the red part above at all...

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?

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

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!

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.

[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).

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:

GMAT assassins aren't born, they're made,
Rich

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

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/

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