# Direct proof that $(5/p)=1$ if $p\equiv 1\pmod{5}$.

I'm trying to show without use of quadratic reciprocity that $(5/p)=1$ if $p\equiv 1\pmod{5}$. If $p\equiv 1\pmod{5}$, then there exists some $x\in U(\mathbb{Z}/p\mathbb{Z})$ with order $5$.

I note that modulo $p$, $(x+x^4)^2+(x+x^4)-1=0$, since $$(x+x^4)^2+(x+x^4)-1=x^2+2x^5+x^8+x+x^4-1=x^4+x^3+x^2+x+1=0.$$

Investigating this relation was the hint in my book, (Ireland and Rosen, question 15 on page 63), but I don't see how it helps. I squared $(x+x^4)^2=1-(x+x^4)$, and got $$(x+x^4)^4=2-3(x+x^4)$$ and $$(x+x^4)^8=13-21(x+x^4)$$ so I think I'm going off the track.

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– Andrés Caicedo Jan 6 '13 at 6:13

The hint in the book implies that the equation $y^2 + y - 1 = 0$ has a root in $\mathbb{Z}/p\mathbb{Z}$. Now use the quadratic formula (which is valid over any field of characteristic $\ne 2$.)
Multiply your relation by $4$. We get $$4(x+x^4)^2+4(x+x^4)-4\equiv 0\pmod{p}.\tag{1}$$ Complete the square: $(1)$ is equivalent to $\left(2(x+x^4)+1\right)^2\equiv 5\pmod{p}$.