4
$\begingroup$

How to prove that the equation $x^2+5=y^3$ has no integer solutions? I have proved the case when $x$ is odd. I used the fact $x^2\equiv 1 \pmod 4$ but how would you do for even $x$: the mod 4 analysis becomes useless. The problem is from Fermat Little Theorem section. But I do not know how apply it. Thanks

  • 3
    You might find a solution [here](https://docs.google.com/viewer?a=v&q=cache:XzxJYDz5GeIJ:www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf+&hl=en&gl=uk&pid=bl&srcid=ADGEEShouNA-L01H_YNGLXaxOQfZ1TpY8LcXKrH7nRaKXqjU90rRnaNQhiH-8qnm5o8T6G_Jl2KblhYZ_Jx4OdtQ2Gpv28aqbq3sOLT2AQz9I74nnGIf1KOOePYHUvXeNDMsq1q6li7l&sig=AHIEtbQWaYmp-5s7ZGYZKll7CzP_BeB71Q) on page 2 - theorem 2.22012-10-29
  • 2
    @OldJohn: your link doesn't seem to work for me (Google docs login...) but I think that it's Conrad's paper available [here](http://ohkawa.cc.it-hiroshima.ac.jp/AoPS.pdf/Examples%20of%20Mordell's%20Equation%20-%20Keith%20Conrad.pdf) too.2012-10-29
  • 0
    Yes - that is the paper. Never really understood the Google docs system for links, I'm afraid.2012-10-29

1 Answers 1

3

Here is a proof. First note that, $y$ cannot be congruent to $3$ modulo $4$. Indeed, had it been; then we would have had $$ x^2 + 5\equiv y^3 \equiv27\pmod{4}\implies x^2 \equiv2\pmod{2}, $$ a contradiction. With this in mind, rewrite the equation as, $$ x^2+4 = y^3 - 1=(y-1)(y^2+y+1). $$ Now, we claim that right hand side always has a prime divisor of form $4k+3$. Indeed, if $y\equiv 0\pmod{4}$, then $y-1\equiv3\pmod{4}$, hence it has such a prime divisor; and if $y\equiv1$ or $y\equiv2$ modulo $4$, $y^2+y+1$ has a prime divisor congruent to $3$ modulo $4$.

Equipped with the following well-known lemma, we are done. If $p\equiv 3\pmod{4} $ a prime, then $$ p\mid x^2+y^2 \implies p\mid x \ \text{and} \ p\mid y. $$ Applying the lemma we conclude for this prime divisor that $p\mid x^2+2^2$ implies $p \mid 2$, contradiction. Done.