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I need to prove that the rank of the curve $y^2=x^3+px$ is $0$, if $p\equiv 7 \pmod {16}$ is a prime. Using the standard technique, we need to show that none of the following two equations admits an integer solution in M, N and e (with M, N and e pairwise coprime; M, e non-zero):
$2M^4-2pe^4=N^2$
$4M^4-pe^4=N^2$

I have got this after going modulo 16. But, $2M^4-2pe^4=N^2$ and $4M^4-pe^4=N^2$ do admit solutions in $Z/16Z$

Shiva
  • 561

2 Answers2

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This is shown in Silverman's "The Arithmetic of Elliptic Curves", Chapter X, Section 6 (The curve $Y^2=X^3+DX$), Proposition 6.2.

Álvaro Lozano-Robledo
  • 15,390
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For $2M^4-2pe^4=N^2$, we see that $N=2n$ for some integer $n$. Substituting in the equation and reducing by 2 we get $2n^2=M^4-pe^4$. Now if $p\equiv 7(mod 16)$ then we get $n^2\equiv -3 (mod 16)$ so $n^2\equiv -3 (mod 8)$ which is a contradiction. Because $x^2\equiv 0,1,4(mod 8) $ for integer $x$.

Sara.Beytasi
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