Suvorov wrote:

Bunuel wrote:

only one such pair is possible \(x=4>y=2\): \(x^y=4^2=16=2^4=y^x\) --> \(x^4-y^4=240\). Sufficient.

How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it.

I think it's worth remembering that \(4^2=16=2^4\), I've seen several GMAT questions on number properties using this (another useful property \(8^2=4^3=2^6=64\)).

But if you don't know this property:

Given: \(x^y = y^x\) and \(x>y\), also \(x\) and \(y\) are

distinct positive integers.

Couple of things:

\(x\) and \(y\) must be either distinct positive odd integers or distinct positive even integers (as odd in ANY positive integer power is odd and even in ANY positive integer power is even).

After testing several options you'll see that \(x=4\) and \(y=2\) is the only possible scenario: because, when \(y\geq{2}\) and \(x>{4}\), then \(x^y\) (bigger value in smaller power) will be always less than \(y^x\) (smaller value in bigger power): \(5^3<3^5\), or \(6^2<2^6\), or \(8^2<2^8\), or \(10^2<2^{10}\), ....

Want to solve a GMAT like problem using this property 8^2=4^3=2^6=64 ??? Do you have any in your arsenal??Pls share