What are the positive integers x for which

$\displaystyle 3^x + 4^x+ ... + (x+2)^x= (x+3)^x$

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- Feb 16th 2008, 09:22 PMperashpositive integers x
What are the positive integers x for which

$\displaystyle 3^x + 4^x+ ... + (x+2)^x= (x+3)^x$ - Feb 16th 2008, 10:33 PMangel.white
Wow, thats a tough question for pre algebra.

I am not good enough to prove it, but the answers are going to be 2 and 3.

$\displaystyle 3^2+4^2 = 9+16 = 25 = 5^2$

$\displaystyle 3^3+4^3+5^3 = 27+64+125 = 216 = 6^3$

After this, the (x+3)^x begins to outpace the sum of the lower powers, so it will continue to grow at a faster rate, and they won't ever catch back up. You can see it by charting the next few but like I said, I'm not good enough to prove it. Maybe someone else can prove it, I would love to see their thought process.