It appears that the
The discriminant of this polynomial is $\pm (-1)^{\binom{p}{2}} 2 p^{p-3} (p+1)^{p-2}$. At leastThus, I have whenever $p$ is of the form $2k^2-1$ with $k$ odd, the discriminant will be a perfect match to numerical data for $p \leq 20$square and the Galois group will be a subgroup of the alternating group. (I didn't restrict myself to primes or to odd $p$.
The data That explains all the counterexamples we have found. We now verify that this is below, in case someone finds it usefulthe discriminant.
I'm finding signs difficult tonight, although it is hard so I'll only get my formulas right up to imagine it being more useful than this formula isa sign.
Here are
We start with the discriminants for following observation. Let $p$ between f$ be a polynomial of degree $1$ and n$. Then $20$ $\mathrm{Disc} ( I saw no reason to limit myself to odd or prime numbers.f(x) This answer is community wiki(x-1)) = f(1)^2 \mathrm{Disc} (f(x)) \quad (*)$$To see this, so feel free to improve itlet $r_1$, $r_2$, .
1, 1037232.., $r_n$ be the roots of the polynomial $f$ and remember that $\mathrm{Disc}(f) = \prod_{i < j} (r_i-r_j)^2.$Now, -157351936, recall the identity$$\mathrm{Disc}(x^{n+1}-ax+b) = \left( (n+1)^{n+1} b^n - 34828517376, 10628820000000, 4287177620000000, n^n a^{n+1} \right).$$It is easy to check that the right hand side vanishes if and only if $x^{n+1}-ax+b$ has a root in common with its derivative $(n+1)x^n-a$, so the two sides agree up to a constant. Checking the constant is left as an exercise.
Plugging in $a=b+1$, we get$$\mathrm{Disc}(x^{n+1}-(b+1)x+b) = \left( (n+1)^{n+1} b^n - 2212089484921012224, n^n (b+1)^{n+1} \right).$$Applying $(*)$ to the identity $x^{n+1}-(b+1)x+b = (x-1)(x^n+x^{n-1}+\cdots+x^2+x-b)$, we get$$\mathrm{Disc}(x^n+x^{n-1}+\cdots+x^2+x-b) = \frac{ (n+1)^{n+1} b^n - 1422639075197644701696,1116533893868237692137472n^n (b+1)^{n+1} }{(n-b)^2}.$$
Taking the limit as $b \to n$, 1050832501626663000000000000and using l'Hospital twice, we get$$\mathrm{Disc}(x^n+x^{n-1}+\cdots+x^2+x-n) =$$$$ (-1)^{\binom{n+1}{2}} \frac{ (n+1)^{n+1} n (n-1) n^{n-2} - 1168651117953810432000000000000, n^n (n+1) n (n+1)^{n-1} }{2}$$$$= \frac{ (n+1)^n n^{n-1} \left( (n+1)(n-1) - 1516612633900744056833610579181568, 2271969367159702514544303572839366656, 3892021112498970434847682120373958672384n^2 \right) }{2} = \frac{ (n+1)^n n^{n-1} }{2}$$
Now, -7561318591827048338056806400000000000000000,
Here they are in factored form. The syntax is that {{2,3}, {5,2}} means applying $2^3 5^2$.
{{{1, 1}}}, {{{1, 1}}}, {{{2, 3}}}, {{{2, 3}, {5, 2}}}, {{{2, 4}, {3, 3}, {5, 2}}}, {{{2, 4}, {3, 3}, {7, 4}}}, {{{2, 16}, {7, 4}}}, {{{2, 16}, {3, 12}}}, {{{2, 8}, {3, 12}, {5, 7}}}, {{{2, 8}, {5, 7}, {11, 8}}}, {{{2, 19}, {3, 9}, {11, 8}}}, {{{2, 19}, {3, 9}, {13, 10}}}, {{{2, 12}, {7, 11}, {13, 10}}}, {{{2, 12}, {3, 12}, {5, 12}, {7, 11}}}, {{{2, 53}, {3, 12}, {5, 12}}}, {{{2, 53}, {17, 14}}}, {{{2, 16}, {3, 30}, {17, 14}}}, {{{2, 16}, {3, 30}, {19, 16}}}, {{{2, 35}, {5, 17}, {19, 16}}}, {{{2, 35}, {3, 18}, {5, 17}(*)$ once more, {7, 18}}we get$$\mathrm{Disc}(x^{n-1} + 2 x^{n-2} + \cdots + (n-1) x + n )$$$$ \frac{ (n+1)^n n^{n-1} }{2} \ \left( \frac{n(n+1)}{2} \right)^{-2} = 2 (n+1)^{n-2} n^{n-3}.$$
