TITLE: Quintic polynomial with Galois Group $A_5$
QUESTION [13 upvotes]: A recent question asks what makes degree 5 special when considering the roots of polynomials with integer coefficients etc. One answer is that the Galois Group of $S_5$ is not solvable. What I am looking for is the most straightforward example (with proof) of a polynomial with integer coefficients and Galois Group $A_5$.
Such an object ought to be standard ... if I ever knew one, I have forgotten it.

REPLY [9 votes]: I'll borrow from DonAntonio's answer the part about the discriminant of the polynomial
$f=X^5+20X+16\in\mathbb Q[X]$: this is $2^{16}5^6$ and thus is a perfect square. Then $G_f$, the Galois group of $f$ over $\mathbb Q$, is contained in $A_5$. 
It remains to prove that $G_f=A_5$. 
First reduce the polynomial modulo $3$: $\overline f=X^5+2X+1\in\mathbb Z_3[X]$, and prove that $\overline f$ is irreducible. Since $G_{\overline f}$ is a subgroup of $G_f$, it follows that $G_f$ contains a $5$-cycle. 
Second, reduce the polynomial modulo $7$: $\hat f=X^5-X+2\in\mathbb Z_7[X]$, and note that $\hat f=(X+2)(X+3)(X^3+2X^2+5X+5)$. Furthermore, $X^3+2X^2+5X+5$ is irreducible over $\mathbb Z_7$ and for the same reason as above $G_f$ contains a $3$-cycle. 
In particular, the order of $G_f$ is divisible by $15$ and then $[A_5:G_f]\leq 4$. On the other side, $A_5$ can't contain a proper subgroup of index less than $5$ (this is true for every nonabelian simple group). It follows that $G_f=A_5$. 
Edit. In this case the simplest way to compute the discriminant is to use the determinant formula which involves the power sums $s_i=x_1^i+\cdots+x_5^i$, where $x_i$ are the roots of $X^5+aX+b$. (This formula can be found in Jacobson, Basic Algebra I, page 258.) After some easy calculations one finds the discriminant: $2^8a^5+5^5b^4$.