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A few hours ago I asked this question. This problem came up while working on a graph labeling problem. I already have a exponential algorithm working. But I want to further reduce the complexity.

Turns out there are too many solutions for $o_1^2-o_2^2=2(o_3^2-o_4^2)$, $o_i$ being odd, such that $o_1>o_2$ and $o_3>o_4$. This doesn't allow me to get a $\mathcal{O}(n^2)$ algorithm for the labeling.

So now I am concentrating on getting a $\mathcal{O}(n^3)$ to work, which requires $o_1^3-o_2^3 = 2(o_3^3-o_4^3)$ ($o_i$ being odd, and $o_1>o_2$ and $o_3>o_4$) to have either no solution or a few solutions. Is this possible??

If not, does a positive integer $x$ exist such that $o_1^x-o_2^x = 2(o_3^x-o_4^x)$ have no (or a few) solutions.

Seems I have to work on my Number theory concepts.

Any hints will be appreciated. Thanks for your patience.

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1 Answers1

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For the equation:

$$x^3-y^3=t(z^3-q^3)$$

You can simply write symmetric solutions.

$$x=at(a^6t^4-3t^2a^3b^3+2b^6)p^2+3at(2b^3-t^2a^3)ps+3tas^2$$

$$y=at(a^6t^4-b^6)p^2-3t^3a^4ps+3tas^2$$

$$z=-b(2t^4a^6-3t^2a^3b^3+b^6)p^2+3b(2t^2a^3-b^3)ps-3bs^2$$

$$q=b(t^4a^6-b^6)p^2-3b^4ps-3bs^2$$

$a,b,p,s$ - integers asked us and any sign.

individ
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