# How do I find p for equations of the form $\sum \limits_i \frac{a_i}{b_i^p} = 1$

The problem I'm facing is solving the following equation for $p$ given the constants $a_i$ and $b_i$:

$$\sum_i \frac{a_i}{b_i^p} = 1$$

Is there a general technique that would allow me to find a value for $p$? If not, are there any heuristic approaches that do not require a computer I can use to find a value for $p$?

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The sum is finite or infinite? the $a_i, b_i$ are real numbers, complex numbers, natural numbers? The exponent $p$ is a prime number, or what? –  GEdgar Oct 25 '11 at 0:38

## 1 Answer

Even a very simple example such as $${1\over2^p}+{1\over3^p}=1$$ is going to give you an equation that can only be solved by numerical techniques. The example $${1\over2^p}+{1\over32^p}=1$$ leads very quickly to the equation $$x^5+x-1=0$$ which again is going to call for numerical techniques. So I think you are asking for a lot.

EDIT: As DSM points out, my 2nd example is a bad one, that quintic actually factors over the rationals. But there are plenty of $S_5$ quintics of the form $ax^5+bx-1$, just take one and then consider $${b\over2^p}+{a\over32^p}=1$$

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You're of course right in general, but I think x^5+x-1=0 is analytically soluble. –  DSM Oct 25 '11 at 1:54
@DSM, I'm not sure I understand your comment. What do you mean by "analytically solvable"? There is no solution in closed form in temrs of powers, exponentials, logs, trig functions, and the four arithmetical operations. –  Gerry Myerson Oct 25 '11 at 2:02
am I simply mistaken? What about x=1/2*Isqrt(3) + 1/2, -1/2*Isqrt(3) + 1/2, (1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3) + 1/3/(1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3), -1/2*(-Isqrt(3) + 1)*(1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3) - 1/6*(Isqrt(3) + 1)/(1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3), and -1/2*(Isqrt(3) + 1)*(1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3) - 1/6*(-Isqrt(3) + 1)/(1/18*sqrt(3)*sqrt(23) + 1/2)^(1/3)? Certainly an arbitrary quintic isn't soluble, but unless I've made a mistake, this one is. –  DSM Oct 25 '11 at 2:20
@DSM, you're absolutely right. I'll edit. –  Gerry Myerson Oct 25 '11 at 3:39
@Gerry Myerson: I always had trouble with minus signs. –  André Nicolas Oct 25 '11 at 9:24