Proof of $n$ being quadratic residue for primes of the form $4n+1$

I'm trying to prove the following statement:

If $$4n+1$$ is a prime $$p$$, then $$n$$ is a quadratic residue $$\bmod p$$.

For this, I thought I could evoke the quadratic reciprocity law and deduce:

$$\genfrac(){}{0}{n}{4n+1}\genfrac(){}{0}{4n+1}{n} =(-1)^{(n-1)\frac{4n+1-1}{4}}=(-1)^{(n-1)n}=1 \\\iff\genfrac(){}{0}{n}{4n+1}=(\genfrac(){}{0}{4n+1}{n})^{-1} =(\genfrac(){}{0}{1}{n})^{-1}=1$$

with Legendre Symbols, but then it occured to me that $$n$$ need not be prime. Looking for a workaround, I found I can deduce that the Jacobi-Symbol must be $$1$$, but If I I can deduce that the Jacobi-Symbol must be $$1$$ but this does not necessarily imply that $$n$$ is a quadratic residue. How do I work around this?

• This might be easier with Euler's criterion. For example, is $4n$ a quadratic residue? Commented Jan 11, 2021 at 16:37
• Yes, then we can invoke multiplicativity-that's elegant. Commented Jan 11, 2021 at 16:38
• Quadratic reciprocity only holds for primes $p$ and $q$. You should probably try another approach altogether. Commented Jan 11, 2021 at 16:38
• The Jacobi symbol $\left(\frac n{4n+1}\right)$ matches the Legendre symbol here because $4n+1$ is prime, if I'm not mistaken. So I think your computation is a proof. Commented Jan 11, 2021 at 16:42
• @IMOPUTFIE The Jacobi symbol and the Legendre symbol are equal when the "denominators" are prime, by definition of the Jacobi symbol. Commented Jan 11, 2021 at 18:02

HINT: $$n$$ is a quadratic residue if and only if $$-1$$ is a quadratic residue.

Indeed, if $$-1=a^2 \pmod p$$, then $$(2^{-1} a)^2 = 4^{-1}a^2 = (4^{-1})(-1) = (4^{-1})(4n) = n \pmod p$$