Let $E/\mathbb{Q}$ the elliptic curve $Y^2=X^3+p^2X$ with $p \equiv 5 \pmod 8$. Show that the abelian group $E(\mathbb{Q})$ has $\text{rank}=0$.
Could you give me a hint how we could do this?
It is known that if $E|_{\mathbb{Q}} Y^2=X^3+aX+b \ \ (a,b \in \mathbb{Z}, D(f) \neq 0)$ then:
$$E(\mathbb{Q})=\{ (\alpha, \beta) \in \mathbb{Q} \times \mathbb{Q} | \beta^2=\alpha^3+a \alpha+ \beta \} \cup \{ [0,1,0]\}$$
EDIT: $p$ is a prime.
This is not true in general. For $p=-3$ we have $p\equiv 5\bmod 8$, but the elliptic curve $y^2=x^3+9x$ has rank $1$ over $\mathbb{Q}$. Also, for positive $p$ this need not be true. Take $p=69$. Then $p\equiv 5 \bmod 8$ and the elliptic curve $y^2=x^3+4761x$ has rank $2$. So we need an additional assumption on $p$. A possible reference for such assumptions is Silverman's "The Arithmetic of Elliptic Curves", Chapter X, Section 6, "The curve $y^2=x^3+Dx$".
Edit: For primes $p$ the question has been asked already on MO and on this site in $2012$, see here and here. However, there seems to be no obvious condition, except the ones in Silverman's book. Perhaps we should ask your professor about the solution.