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If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?

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  • $\begingroup$ what if $p=\sqrt{2}$ or is an irrational $\endgroup$ – user9413 May 17 '11 at 13:27
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    $\begingroup$ Not without more knowledge about $a, b, c$, or $p$. All that condition tells you is that, say, $a = \sqrt{c} \cos \theta, b = \sqrt{c} \sin \theta$ for some $\theta$, which doesn't uniquely determine the value of $a^p + b^p$ for (I think) any $p \neq 2$. $\endgroup$ – Qiaochu Yuan May 17 '11 at 13:30
  • $\begingroup$ @Chandru: If $p=\sqrt{2}$ has a solution, it would be great to discuss it here as well. $\endgroup$ – Ahmed May 17 '11 at 13:31
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    $\begingroup$ A simple example to illustrate why this cannot work: if $a$ and $b$ are solutions, then so are $-a$ and $-b$. But if $p$ is an odd integer, then this gives different values of $a^p + b^p$. $\endgroup$ – Alex B. May 17 '11 at 13:35
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The other responses showing you cannot find it just from $a^2+b^2=c$ are correct. You may be interested in the fact that given $a^2+b^2=c, a+b=d$, you can form $(a^2+b^2)(a+b)=cd=a^3+a^2b+ab^2+b^3$ and $d^2-c=2ab$, so $cd-\frac{d^3-cd}{2}=a^3+b^3$. Given as many symmetric polynomials as variables you can find all the higher orders.

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There are $a, b, A, B$ such that $a^2+b^2=A^2+B^2$ but $a^p+b^p \ne A^p+B^p$. So the answer to your first question is NO.

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  • $\begingroup$ For example $1^2+8^2=65$ and $4^2+7^2=65$, but $1^3+8^3=513$ and $4^3+7^3=407$ $\endgroup$ – Henry May 17 '11 at 13:53
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Since $1^2 + 7^2 = 5^2 + 5^2$, the number $c$ does not uniquely describe $a$ and $b$. Therefore, you can get not find $a^p + b^p$.

See Numbers which are the sum of two squares in two or more different ways for a list of more such examples.

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  • $\begingroup$ Not exactly relevant, since the question is about real numbers. The list Real numbers which are the sum of two squares of real numbers in two or more different ways is rather uninteresting. $\endgroup$ – GEdgar May 17 '11 at 15:22
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There are times where $c$ does uniquely specify $a,b$, which is maybe worth mentioning. Suppose $a,b,c$ are all positive integers. Then we can look at the sum of squares function for $c$, and we know that (up to order) $a,b$ will be unique when $c$ has exactly one prime factor of the form $4k+1$, and each prime factor of the form $4k+3$ appears to an even power.

In other words, if $c$ is of that form, and $a^2+b^2=c$, $a>0$, $b>0$, then $a^p+b^p$ is completely determined for any $p$.

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