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

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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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New post 01 Sep 2015, 04:31
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anik19890 wrote:
Bunuel wrote:
axl_oz wrote:
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 \(x^2+y^2\).

how to simplify this withlout calculator 2^x = x^2-1


You can simplify via :

1. Graphical method. Plot graphs for \(2^x\) and \(x^2-1\) and see where these intersect and these points of intersections will give you possible values of x satisfying the given equation or

2. Iterative process wherein you find what value of x satisfy the equation \(2^x\) = \(x^2 - 1\) , in this case as \(2^x\) > 0 (for all x), I will not use x<0 values, x=3 satisfies the value.

Graphical method is a bit more 'complex' if you are not comfortable with graphs and coordinate geometry. If given as a part of GMAT question, rest assured you can follow method 2 and you will be able to find a small enough value satisfying the given equation. In the case above, I saw that \(2^x\)> 0 for all x, thus started with x =1,2,3... etc

Hope this helps.

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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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New post 01 Sep 2015, 10:05
[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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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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New post 05 Sep 2015, 19:23
Bunuel wrote:
axl_oz wrote:
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 \(x^2+y^2\).

how did you simplify this 2^x=(x−1)(x+1) --> x=3

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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New post 11 Sep 2015, 02:10
In order to maximize |x-y| we need to have y=0 as y is non negative
this gives us hint and we can back solve by putting Y=0

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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New post 13 Nov 2015, 04:41
Ok but if you set x=4 and y=1 (|x-y| = |4-1| = 3) and use 2 as the base you will have:

2^4+2^1 = 4^2+1^2 ---> 18=17

So the solution works only with y=o ...

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte   [#permalink] 29 Nov 2017, 01:11

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