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I need a proof that this equation has more than one solution for $p$ and $q$. $$p^{q-2}= 1024$$, where $q\in N, q>2$

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closed as off-topic by Jonas Meyer, Ittay Weiss, Meelo, Najib Idrissi, dustin Jan 18 at 18:36

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2  
Explain the down vote someone, this is not how we greet new users. –  Patrick Da Silva Nov 10 '11 at 19:25
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@Patrick: Well, I didn't vote, but it might have to do with the attitude displayed in the previous questions and the user profile... –  t.b. Nov 10 '11 at 19:28
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@Patrick: This is not a new user. –  Asaf Karagila Nov 10 '11 at 19:28
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How about $q=17$, $p=\sqrt[15]{1024}$, respectively $q=22$, $p=\sqrt{2}$? –  Henning Makholm Nov 10 '11 at 19:31
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@Chun-Yue In that case, you can construct infinitely-many solutions using Makholm's example as a guide. –  Austin Mohr Nov 10 '11 at 19:40

3 Answers 3

Here's a proof:

$(p,q) = (32,4)$ solves the equation, as does $(4,7)$.

QED.

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It seems that everybody so far missed the solution $p=1024$, $q=3$. :-)

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1  
Lol. I'd love to +5 this one. –  Patrick Da Silva Nov 10 '11 at 22:43

How about $2^{10}$ and $4^5$? that gives you $p=2$ and $q=12$, or $p=4$ and $q = 7$. The easiest way to see this is just to factor $1024$ and see the possibilities...

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