Suppose that $a, b$, and $p$ are integers such that $b \equiv 1 \; \pmod{4}$, $p \equiv 3 \; \pmod{4}$, $p$ is prime, and if $q$ is any prime divisor of $a$ such that $q \equiv 3 \; \pmod{4}$, then $q^{p}\vert a^{2}$ and $p$ does not divide $q-1$ (if $q=p$, then also $q \vert b$). Show that the equation \[x^{2}+4a^{2}= y^{p}-b^{p}\] has no solutions in integers.
Problem
Source:
Tags: modular arithmetic, Diophantine Equations
TomciO
25.06.2008 03:20
Conditions of this problem are a little bit weird. Note that $ y \equiv 1 \pmod{4}$, since otherwise the right hand side would be congruent to $ 2$ or $ 3$ mod $ 4$ while the left is congruent to $ 0$ or $ 1$. Let us rewrite the equation in the form: \[ x^2 + 4a^2 = (y-b)\frac{y^p-b^p}{y-b} = (y-b)(y^{p-1} + y^{p-2}b + ... + yb^{p-2} + b^{p-1}).\] Since $ y \equiv b \equiv 1 \pmod{4}$ we obtain that $ \frac{y^p-b^p}{y-b} \equiv 1+1...+1 \equiv p \equiv 3 \pmod{4}$. Let $ d = (y,b)$ and write $ y = dy_0$, $ b=db_0$. Then $ \frac{y^p-b^p}{y-b} = d^{p-1} \frac{y_0^p-b_0^p}{y_0-b_0}$ and since $ d^{p-1} \equiv 1 \pmod{4}$ we conclude that $ \frac{y_0^p-b_0^p}{y_0-b_0} \equiv 3 \pmod{4}$. Therefore this number has a prime divisor of the form $ 4k+3$, call it $ q$. Since $ y_0, b_0$ are coprime $ q$ is not a divisor of $ y_0, b_0$ and we can easily prove that order of $ \frac{y_0}{b_0}$ mod $ q$ is $ p$, so $ p|q-1$. But at the same time we have $ q|x^2 + (2a)^2$ and since $ q=4k+3$ it is well known that we must have $ q|x$ and $ q|a$. That's a contradiction with the given condition. I haven't used all conditions, is it wrong?