Diophantine equations, factoring, modular arithmatic.

Asked Nov 27 '20 at 04:10

Active Nov 27 '20 at 04:10

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Find all integers $n$ such that $x^3 + y^3 + z^3 -3xyz = n$ is solvable in positive integers.

I have made the following observations;

Use identity and rewrite the expression as $\frac{1}{2}(x+y+z)$ $([x-y]^2 + [y-z]^2 + [z-x]^2)$

but i am not sure how that helps.

Also i noticed $ n \equiv x+y+z (mod 3)$.

I think there'll be infinite $n$ and we need to essentially find (m,r) such that n = mk + r

Any help is appreciated, cheers!

asked Nov 27 '20 at 04:10

Aditya_math

5

Possible duplicate of Integers can be expressed as $a^3+b^3+c^3-3abc$ – Parcly Taxel Nov 27 '20 at 04:12
@ParclyTaxel i dont understand why they have taken (x,y,z) = (k,k,k+1) , i mean can't they be of form (k+4, k+6, k+9)? – Aditya_math Nov 27 '20 at 04:54
$(k,k,k+1)$ is enough to show that all integers of a certain form may be achieved, and similarly for the other case. That's all we need here. – Parcly Taxel Nov 27 '20 at 04:57
@ParclyTaxel are they talking about something like k,k,k+1 mod 3. I am not able to understand it well. – Aditya_math Nov 27 '20 at 05:03
There are always solutions. https://math.stackexchange.com/questions/3351039/cubic-diophantine-equation-with-a-prime-x3-y3-z3-3xyz-p/3352441#3352441 – individ Nov 27 '20 at 05:44

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