Three diophantine equations

by #H34N1, Jan 2, 2009, 2:25 PM

These ones are based off of Engel-like NT problems:

1. Find all positive integral solutions $ (x,y,z,k)$ to $ x^3 + y^3 + z^3 = kxyz$.
2. Find all positive integral solutions $ (x,y,z,k)$ to $ x^3 + y^3 + z^3 = x^2y^2z^2$.
3. Find all positive integral solutions $ (x,y,z,k)$ to $ x^3 + y^3 + z^3 = kx^2y^2z^2$.

meh I lost
diophantine.equations
nt

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quite a coincidence. I was just flipping through the NT section in Engel. Pell-Fermat? idk. i'm too n00bish

by Poincare, Jan 2, 2009, 5:05 PM

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Uh how would you use a pells equation?

by #H34N1, Jan 3, 2009, 3:07 AM

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Where's the k in question 2?

by andersonw, Jan 12, 2009, 2:20 AM

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Ok. I got some dumb idea on Q3.
$k$ can be $1$ or $3$. WLOG,assume that $x\ge y\ge z$. Then we have$3x^3\ge x^3+y^3+z^3=kx^2y^2z^2,$ which gives $x\ge\frac{k}{3}y^2z^2$.
Next, we rewrite the equation as $y^3+z^3=x^2(ky^2z^2x)$. Since the left-hand side is positive, we have $ky^2z^2-x\ge 1$. Thus, $2y^3\ge y^3+z^3\ge(y^2z^2)(1)= y^4z^4$.
This yields $18\ge k^2yz^4$. There is no positive integer solution when $k\ge 5$. When $k=4$, we must have $y=z=1$. The given equation becomes $x^3+2=4x^2$. This shows $x^2\mid 2$, and hence $x=1$. But this is not a solution. When $k=2$, $yz^2\le 4$, so $z=1$ and $1\le y\le 4$. Checking routinely, none of these are solutions.

by Bryan0224, Nov 15, 2024, 3:52 AM

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I just checked $k=1$ and $k=3$, and the only solutions were $(x,y,z,k)=(1,1,1,3)\text{ and }(1,2,3,1)$. That automatically solves Q2 as well.

by Bryan0224, Nov 17, 2024, 1:12 PM

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