Can Nonlinear Congruences Be Solved? A Case Study with a Prime Modulus

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Homework Statement



Find an integer x that is a solution (only need one solution, not all solutions). If no solution exists, prove that no solution can exist.

13x = 13 mod 50032 , with x > 1. Note that 5003 is prime.

Here, = means "congruent to"

Homework Equations



Not sure. I can solve linear congruences without too much trouble, but I haven't seen a congruence of this form before, and the book that I'm using is pretty low on good examples.

The Attempt at a Solution



I don't have one. Looking for a bump in the right direction.
 
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I spent a lot of time last night trying a bunch of different techniques on your problem and couldn't get much of anywhere. But since no one else has responded to it yet I will give you what little I have, which is just Fermat's Little Theorem:

13^5003 = 13 mod 5003. But from here I don't know where to go.
 
I can use Euler's theorem since 13 and 50032 are coprime.

However, I then end up with: 13phi(50032) = 1 mod 50032. How can I manipulate this so that I have 13 mod 50032 instead of 1 mod 50032?
 
Can't you just multiply both sides by 13?
 
Sigh ... yes. Yes I can. I think it's time for a break, I've been staring at this for too long and I'm starting to forget about basic math.

Thanks for the wake up call!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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