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. 

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 $\left(2(x+x^4)+1\right)^2\equiv 5\pmod{p}$.
