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let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$

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  • $\begingroup$ There is one particular solution: math.stackexchange.com/questions/784612/another-triple $\endgroup$ – individ Nov 8 '14 at 4:34
  • $\begingroup$ In General, for such equations the solution is very cumbersome. For example: math.stackexchange.com/questions/772409/… Why do you need such a solution? These formulas are always removed. What are you going to do with them? $\endgroup$ – individ Nov 8 '14 at 5:42
  • $\begingroup$ @individ, for the millionth time NO FORMULAS please... give me a method...Thanks. $\endgroup$ – user97615 Nov 8 '14 at 12:38
  • $\begingroup$ Have you tried any method which did not work due to some difficulties? Maybe posting some efforts can help others find a valid method? $\endgroup$ – awllower Nov 10 '14 at 4:59
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It is a theorem that one can identically solve,

$$x_1^2+x_2^2+\dots+x_n^2 = (y_1^2+y_2^2+\dots+y_n^2)^k$$

for any positive integer n and k. Thus the kth power of n squares is itself the sum of n squares. For example, for $n =3$, we have,

$k=2:$

$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2 = (a^2 + b^2 + c^2)^2$$

$k=3:$

$$a^2(a^2 - 3b^2 - 3c^2)^2 + b^2(-3a^2 + b^2 + c^2)^2 + c^2(-3a^2 + b^2 + c^2)^2=(a^2 + b^2 + c^2)^3$$

and so on. See Theorem 1 at https://sites.google.com/site/tpiezas/004.

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  • $\begingroup$ Thank you so much @Tito Piezas III $\endgroup$ – user97615 Nov 10 '14 at 19:27
  • $\begingroup$ @NumThcurious You're welcome. Did you see my answer to your question regarding Lebesgue's Identity? $\endgroup$ – Tito Piezas III Nov 10 '14 at 20:13
  • $\begingroup$ Not yet. let me check it out. Thanks once again. $\endgroup$ – user97615 Nov 10 '14 at 20:48
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For the equation:

$$x^2+y^2+z^2=r^3$$

Will make a replacement that formula was compact.

$$c=2(q-p-s)t$$

$$d=s^2+t^2-q^2-p^2+2p(q-s)$$

$$k=p^2+t^2-q^2-s^2+2s(q-p)$$

$$n=p^2+t^2+s^2-q^2$$

$$j=p^2+s^2+t^2+q^2-2q(p+s)$$

$p,s,t,q$ - integers asked us. Then decisions can be recorded.

$$x=dn^2+2cnj-dj^2$$

$$y=cj^2+2dnj-cn^2$$

$$z=k(n^2+j^2)$$

$$r=n^2+j^2$$

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  • $\begingroup$ I really appreciate you are taking the time to write down all that, which I cannot check whether it's accurate or not. But honestly,you are driving me crazy with your formulas: I need a stepwise approach... a method. Please NO FORMULAS. Show me HOW you did it. If you don't know, please say so. $\endgroup$ – user97615 Nov 9 '14 at 12:25
  • $\begingroup$ @NumThcurious The method does not publish. To use prohibited. It is constantly trying to erase. The forum is not about to tell. Don't want to lose priority. $\endgroup$ – individ Nov 9 '14 at 12:59
  • $\begingroup$ @individ Why your comment above sounds like the result of Google translate? Per chance you are not a fluent native English user? (Just out of curiosity, no offense intended.) $\endgroup$ – awllower Nov 10 '14 at 4:57
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    $\begingroup$ @awllower I know only Russian language. If writing is not clear, excuse me. So I try to only write the formula. $\endgroup$ – individ Nov 10 '14 at 5:03
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    $\begingroup$ @awllower thank you. But now the longer I will not be able to understand the better for me. Better help to open the topic for systems of nonlinear Diophantine equations. There was this, but deleted it. To complain of steel that many formulas wrote. This system I decided. mathoverflow.net/questions/146768/… $\endgroup$ – individ Nov 10 '14 at 5:15
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The simple formula for the equation:

$$x^2+y^2+z^2=q^3$$

You can write this:

$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$

$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$q=3(p^2+2k^2-2kt+2t^2)s^2$$

$p,k,t$ - integers asked us.

If there is a one simple solution, it will be necessary to reduce to the corresponding $s$.

There are other formulas. But they are bulky. Don't know whether it makes sense to write them.

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  • $\begingroup$ $x,y,z$ are not cop rime in your formula. $\endgroup$ – user97615 Nov 8 '14 at 18:10
  • $\begingroup$ @NumThcurious Did not understand. What is there not to like? $\endgroup$ – individ Nov 8 '14 at 18:13
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    $\begingroup$ i like everything except the fact that $x,y,z,q$ have a common factor. i did mention that $\gcd(a,b,c,d)=1$ $\endgroup$ – user97615 Nov 8 '14 at 18:21

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