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SNFS Polynomial
I do not have access to computer algebra software [e.g. Maple]
2,2190L is divisible by 2,730L. It is easy to divide out this algebraic factor and get a good polynomial. But the number is also divisible by 2,438M. Can someone with a CAS see if one can get a decent SNFS polynomial after dividing out both algebraic factors? |
[QUOTE=R.D. Silverman;103489]I do not have access to computer algebra software [e.g. Maple]
2,2190L is divisible by 2,730L. It is easy to divide out this algebraic factor and get a good polynomial. But the number is also divisible by 2,438M. Can someone with a CAS see if one can get a decent SNFS polynomial after dividing out both algebraic factors?[/QUOTE] 2,2370L has the same structure. --> 2370 is divisible by 3 and 5... |
[QUOTE=R.D. Silverman;103491]2,2370L has the same structure. --> 2370 is divisible by 3 and 5...[/QUOTE]
And while we are at it.... 2226 is divisible by 3 and 7, so can we get a good polynomial for 2,2226L,M? |
[QUOTE=R.D. Silverman;103489]I do not have access to computer algebra software [e.g. Maple]
2,2190L is divisible by 2,730L. It is easy to divide out this algebraic factor and get a good polynomial. But the number is also divisible by 2,438M. Can someone with a CAS see if one can get a decent SNFS polynomial after dividing out both algebraic factors?[/QUOTE] Don't think so. At least I don't see a clever way of making something useful out of 1 + 2x - x^2 - 50x^3 + 93x^4 - 46x^5 - 2x^6 + 6x^7 - x^8 - 3x^37 + 5x^38 + 20x^39 - 31x^40 - 8x^41 + 23x^42 - 9x^43 + x^44 + x^73 + 2x^74 - 23x^75 + 27x^76 + 4x^77 - 12x^78 + 3x^79 - 3x^110 + 15x^111 - 9x^112 - 17x^113 + 15x^114 - 3x^115 - x^147 - 13x^148 + 31x^149 - 18x^150 + 3x^151 + 9x^184 - 15x^185 + x^186 + 4x^187 - x^188 - x^219 - 2x^220 + 2x^221 + 8x^222 - 6x^223 + x^224 + 2x^256 - 2x^257 - 4x^258 + 2x^259 - x^292 + 5x^294 - 2x^295 + 2x^329 - 6x^330 + 2x^331 - x^365 + 2x^366 + 2x^367 - x^368 - 3x^403 + x^404 + x^439 - x^475 + x^511 - x^548 + x^584 Greg |
And likewise for the other cases. Nothing useful after dividing out the second factor.
Greg |
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