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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink]
1
motion2020 wrote:
given question's stem, compare Quantity A with Quantity B for 4 choices greater/less/equal/different always

\(2^x+2^y+2xy=(x+y)^2\)
this is possible when x=y=2 ---> \(4+4+2*2*2=(2+2)^2=16\) making Quantity A < Quantity B
and, when x=4, y=2 ---> \(16+4+2*4*2=(4+2)^2=36\) making Quantity A = Quantity B
Hence, answer must be D
KarunMendiratta wrote:
SUPER HARD

\(2^x + 2^y = x^2 + y^2\), where \(x\) and \(y\) are non-negative integers.

Quantity A
Quantity B
Greatest possible value of \(|x - y|\)
2


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


I think you are missing out on - Greatest possible value!
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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink]
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink]
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