Quintic polynomial with Galois Group $A_5$ 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.
 A: You may want to read chapter $\,4\,$ here , page $\,84\,$ in the "ereaders" version.
After some rather messy and annoying calculations, the discriminant of for example the polynomial $\,x^5+20x+16\in\Bbb Q[x]\,$ , which appears in the other answer, is $\,5^2\cdot 80^4\,$ , which is a square in $\,\Bbb Q\,$ and thus its Galois Group is contained in $\,A_5\,$...
A: Here are some found by computer search
$$x^5 - 55x - 88$$
$$x^5 - 55x + 88$$
$$x^5 + 20x - 16$$
$$x^5 + 20x + 16$$
$$x^5 + 95x - 76$$
$$x^5 + 95x + 76$$

$$x^5 + 3x^3 + 5x - 10$$
$$x^5 + 3x^3 + 5x + 10$$
$$x^5 + 6x^3 - 7x - 8$$
$$x^5 + 6x^3 - 7x + 8$$
$$x^5 + 10x^3 - 10x - 4$$
$$x^5 + 10x^3 - 10x + 4$$

$$x^5 - x^4 + x^3 + 2x^2 + x - 1$$
$$x^5 + x^4 + x^3 - 2x^2 + x + 1$$
A: 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: $2^8a^5+5^5b^4$. 
