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## Main Question or Discussion Point

I want to find three integers, a b and c, with a^3 + b^3 = 17 * c^3.

all each the smallest integer possible

all each the smallest integer possible

- Thread starter IndustriaL
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I want to find three integers, a b and c, with a^3 + b^3 = 17 * c^3.

all each the smallest integer possible

all each the smallest integer possible

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matt grime

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saltydog

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Zurtex

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matt grime

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there are algebraic ways of factoring such things in number fields, though i don'y know much about them - they are of elementary (in the sense of important) interest in algebraic number theory.

you can also get conditions on a,b and c by considering it modulo some inetger. eg doing so mod 8 the cubic remainders are 0,1,3,5,7 and we need

a^3+b^3=c^3 mod 8

so that the only non trivisl way to do it is to have a^3=1,b^3=7 mod 8 or a^3=3, b^3=5 mod 8 or a^3=b^3=c^3=0 mod 8

you can also get conditions on a,b and c by considering it modulo some inetger. eg doing so mod 8 the cubic remainders are 0,1,3,5,7 and we need

a^3+b^3=c^3 mod 8

so that the only non trivisl way to do it is to have a^3=1,b^3=7 mod 8 or a^3=3, b^3=5 mod 8 or a^3=b^3=c^3=0 mod 8

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saltydog

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Why not Zurtex? I respect you and Matt's opinions but I'm stubborn. You mind if I check?Zurtex said:

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matt grime

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i meant to say, that the case a^3=b^3=c^3 can be dismissed as trivial since it implies that all of a,b,c are even thus it is not a "primitive" solution, but i forgot the cases

a^3=0 and b^3=c^3 mod 8

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matt grime

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looking mod 7 i think we find that 7 divides c and a^3 and b^3 are +/-1 mod 7 and mod 17 that a+b=0

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saltydog

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Alright, I'd like to qualify Zurtx's statement if I may: Mathematica cannot find any value of a and b under 5000 which satisfy the equation. Frankly, if I had access to a faster PC I'd run it up to a million at least as well as optimize my algorithm. It kinds looks like it's related to Fermat's theorem. Is there a proof that there is no solution?

Wait, he did say integers. Need to check some negative numbers too. Didn't say I was good in Number Theory, only that I'm interested in it.

Wait, he did say integers. Need to check some negative numbers too. Didn't say I was good in Number Theory, only that I'm interested in it.

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Zurtex

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Actually I just used the FindInstance function for c > 0 and then for c < 0.saltydog said:Alright, I'd like to qualify Zurtx's statement if I may: Mathematica cannot find any value of a and b under 5000 which satisfy the equation. Frankly, if I had access to a faster PC I'd run it up to a million at least as well as optimize my algorithm. It kinds looks like it's related to Fermat's theorem. Is there a proof that there is no solution?

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saltydog

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FindInstance? Suppose that's a 5.0 feature. I have 4.1.Zurtex said:Actually I just used the FindInstance function for c > 0 and then for c < 0.

IndustriaL, where you got this anyway? I'd like to see some closure in the matter. You know Richard Guy has a nice book out, "Unsolved Problems in Number Theory". Don't suppose anything like this is in there?

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matt grime

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saltydog

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Very well Matt. I'll leave it there.matt grime said:

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matt grime

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