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I recently discovered that you can get all the primitive pythagorean triples with this neat formula known as Euclid's formula $$a = m^2 - n^2, b = 2mn, c = m^2 + n^2$$ Where, m and n are co-prime and each having a different parity, that is one is odd and the other is even. PrimitivePythagoreanTriples

I am curious to know if it is possible to find such a formula for cubes that is an expression with 2 numbers or multiple numbers $(x_1, x_2, x_3, ...)$ such that setting $ a, b, c, d $ respectively with the expressions satisfies $a^3 + b^3 + c^3 = d^3$ if it does exist can such expressions exist for any integral power n, such that $ a_1^n + a_2^n + a_3^n + ... + a_n^n = b^n$.

Jeeth2006
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  • About pythagorean triples, you may be interested by this article : https://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples – Jean Marie Jan 05 '22 at 10:39
  • Do you the story of Ramanujan about 1729 ? – Jean Marie Jan 05 '22 at 13:27
  • https://math.stackexchange.com/questions/3679773/diophantine-cubic-equations/3680119#3680119 https://math.stackexchange.com/questions/107570/generalization-of-pythagorean-triples/721310#721310 – individ Jan 05 '22 at 15:41

3 Answers3

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Ramanujan found certain parametrizations for the cubic case, for example $$ (3u^2+5uv-5v^2)^3+(4u^2-4uv+6v^2)^3+(5u^2-5uv-3v^2)^3=(6u^2-4uv+4v^2)^3. $$ However, it is known that not all primitive solutions arise this way. So the quadratic case cannot be generalised in such a way.

A very nice report of the Diophantine equation $x^3+y^3+z^3=w^3$ is given here, with many examples.

Dietrich Burde
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The fourth power case is the Jacobi Madden equation: $a^4+b^4+c^4+d^4=(a+b+c+d)^4$ Jacobi Madden equation.

It gives infinitely many ineteger solutions for $a^4+b^4+c^4+d^4=e^4$ using elliptic curve.
Unfortunately, the parametric solution is not known.
Some small solutions are

$955^4 + 1770^4 + (-2634)^4 + 5400^4 = 5491^4$

$48150^4 + (-31764)^4 + 27385^4 + 7590^4 = 51361^4$

$561760^4 + 1493309^4 + 3597130^4 + (-1953890)^4 = 3698309^4$

For $n=5,$ see Diophantine equation 5th power.

Sastry found $(75v^5-u^5)^5+(u^5+25v^5)^5+(u^5-25v^5)^5+(10u^3v^2)^5+(50uv^4)^5 =(u^5+75v^5)^5$

Other parametric solutions are here.

Tomita
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It's essentially the same to ask for merely rational-number solutions, and then clear the denominators if we want.

In that context, the fact that (the projective curve) $x^2+y^2=z^2$ is genus $0$ (while higher-degree analogues are not) means that we can parametrize almost all of it by rational functions of a single numerical parameter, e.g., $x=2t, y=t^2-1, z=t^2+1$. A physical way to find this parametrization is by looking at straight lines in the $x,y$-plane intersecting the circle $x^2+y^2=1$, and passing through the point $(x,y)=(0,1)$, for example.

An essential point is that every straight line intersects that circle in exactly two points (possibly counting multiplicities)...

As Joe Lipman pointed out to me long ago, for $p>2$, Fermat equations $x^p+y^p=z^p$ have no solutions in rational functions $x,y,z$ of an affine parameter $t$, because the curve is of genus $>0$, and such a parametrization would give a (non-trivial) map from a genus-zero curve to a positive-genus curve. This is not possible, by Riemann-Hurwitz, etc. :)

With higher-degree, and more-variables examples, everything falls apart... :)

paul garrett
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