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Mar 15, 2012 at 12:57 vote accept Richard
Mar 15, 2012 at 10:30 comment added Geoff Robinson It might be easier to set $y = x+3$ and work with the polynomial $y^4 - 40y^2 + 120y -80.$
Mar 15, 2012 at 6:31 answer added Zack Wolske timeline score: 1
Mar 15, 2012 at 6:00 answer added Noam D. Elkies timeline score: 5
Mar 15, 2012 at 5:13 comment added Franz Lemmermeyer This was also posted on math stackexchange.
Mar 15, 2012 at 0:43 answer added Aaron Meyerowitz timeline score: 2
Mar 14, 2012 at 23:10 comment added Zack Wolske It seems the congruences you want are $p \equiv \pm 11 \mod 30$, and $p \equiv \pm 1 \mod 30$. The other cases $(7,13,17,23)$ are taken care of.
Mar 14, 2012 at 23:02 comment added Zack Wolske $2$ is a root mod $29$: $16 + 96 + 56 - 24 + 1 = 88 + 57 = 87 + 58 = 3(29) + 2(29)$, so $h$ has a linear factor.
Mar 14, 2012 at 21:43 answer added Will Jagy timeline score: 4
Mar 14, 2012 at 19:18 comment added Richard The extension determined by h over Q is Galois, and SAGE computes it as C4, the cyclic group of order 4, not as the Klein 4 group. By the way, this isn't from a homework question. I'm using it to determine when a parameterized family of elliptic curves has j-invariant 0.
Mar 14, 2012 at 18:52 answer added user19475 timeline score: 2
Mar 14, 2012 at 18:47 comment added user19475 MAGMA calculated the Klein four group $\mathbf{Z}/2 \times \mathbf{Z}^2$. Now use class field theory and Chebotarev.
Mar 14, 2012 at 18:26 comment added user19475 Can you tell us the Galois group of $h$ over $\mathbf{Q}$? It must be abelian if the splitting is determined by congruence conditions.
Mar 14, 2012 at 18:23 comment added Igor Rivin This has a homework scent to it...
Mar 14, 2012 at 18:20 history asked Richard CC BY-SA 3.0