Are there natural number solutions to the equation $a^2(m^3n-n^3m)=b^2(u^3v-v^3u)=c^2(p^3q-q^3p)$ such that the following conditions are met?

Question

Are there solutions to the equation $$a^2(m^3n-n^3m)=b^2(u^3v-v^3u)=c^2(p^3q-q^3p)$$ such that, $$m\neq n$$ $$u\neq v$$ $$p\neq q$$ $$\{m,n\}\neq \{u,v\}$$ $$\{u,v\}\neq \{p,q\}$$ $$\{p,q\}\neq \{m,n\}$$ $$a,m,n,b,u,v,c,p,q\in \mathbb N$$ Are such solutions possible? If possible, what are the examples and how could we generate them?

Answer 1

Yes there are possible solutions and you can easily generate them. If you multiply out one part you get $$ a^2(m^3n-n^3m) = a^2mn (m+n)(m-n) $$ So in fact you are searching for a big number that has some prime factors and you try to find different representations for it. Best is some number with a high power of a prime and we just need to distribute it between all the available variables.

Let's start a bit easier and let's try to find a number that can be represented as above using two notations.

We choose $n=v=4$ and $m=8$ and $u=12$. This already gives us

$$ a^2 \cdot 8 \cdot 4 \cdot (12 \cdot 4) \overset{!}{=} b^2\cdot 12 \cdot 4 \cdot (16 \cdot 8) \\ \Leftrightarrow a^2 = 4b^2 $$ so if we choose $a=8$ and $b=4$ we have found a solution. We can also try to find another solution with $q=8,p=16$, this would give $$a^2 \cdot 8 \cdot 4 \cdot (12 \cdot 4) \overset{!}{=} c^2\cdot 16 \cdot 8 \cdot (24 \cdot 8) \\ \Leftrightarrow a^2 = 16c^2 $$ so with $c=2$ we have found another solution and all of those numbers are pairwise different.

I think that you can actually find infinite solutions. Start with powers of a prime number so that the resulting total number will be close to a power of this prime number and then you might find new solutions by multiplying or dividing by your prime number. In the above example we had the three solutions similar to $(a, n, m)$: $$ (8, 4, 8), (4, 4, 12), (2, 8, 16) $$

This is no general rule but maybe this idea helps to find a more general solution how one could generate those examples.

Answer 2

You asked

Are there solutions to the equation $$a^2(m^3n-n^3m)=b^2(u^3v-v^3u)=c^2(p^3q-q^3p)$$

subject to easy restrictions. This is related to finding Congruent Numbers.

Define a function on ordered pairs with

$$ f(n,m) := m^3 n - n^3 m = n\,m\,(m-n)(m+n). $$

Define an equivalence relation $R$ on pairs of natural (or rational) numbers not both zero such that $(m,n)$ is $R$ equivalent to $(u,v)$ if and only if

$$ a^2(m^3n-n^3m)=b^2(u^3v-v^3u)$$

for nonzero natural numbers $a,b$.

In other words, $(m,n)$ is $R$ equivalent to $(u,v)$ if and only if

$$ f(n,m) = (b/a)^2 f(u,v). $$

More simply stated is that

$$ \text{core}(f(n,m)) = \text{core}(f(u,v)) $$

where $\,\text{core}(n)\,$ is the squarefree part of $\,n\,$ (OEIS sequence A007913).

Based on the congruent number $6$, a quick search for pairs such that $\, f(n,m) = 6k^2\,$ finds the pairs

$$(1,2),\,(1,3),\,(1,49),\,(24,25),\,(529,2738),\,(2209,3267).$$

Because they all satisfy $\, f(n,m) = 6k^2,\,$ then they are all $R$ equivalent to each other. Pick any three of them and they will satisfy the given equation.

Define an equivalence relation $E$ on pairs of natural (or rational) numbers not both zero such that $(m,n)$ is $E$ equivalent to $(u,v)$ if and only if $(am, an) = (bu, bv)$ for nonzero natural numbers $a,b$. The previous pairs are all mutually not $E$ equivalent.

Similarly with the congruent number $5$, a quick search for pairs such that $\, f(n,m) = 5k^2\,$ finds the pairs

$$(1,9),\,(4,5),\,(720,1681),\,(961,2401).$$

They are all $R$ equivalent to each other but not $E$ equivalent. How to generate such pairs?

One answer uses generalized Somos-4 sequences as used in my "WXYZ Math Project". For the specific case of congruent number $5$, this is directly connected to the four OEIS sequences A129206 to A129209.

For example, for the index $2$, the four sequence values are $\,-12,\,-49,\,41,\,31.$ Notice that

$$(720,1681) = (5\,(-12)^2, 41^2)\quad \text{ and }\quad (961,2401) = (31^2,(-49)^2) $$

and that this generalizes for any index in the four sequences. All four sequences satisfy the common recursion

$$ a_n= (\pm 144\,a_{n-1}a_{n-3} + 2257\,a_{n-2}^2 ) / a_{n-4}. $$

with different initial values and with the sign of $144$ being $+$ for the first two sequences and $-$ for the other two sequences.

For example, for the sequence A129209 the first few sequence terms are

$$ a_0 = 1,\, a_1 = 3,\, a_2 = 31,\, a_3 = 5283,\;\; \text{ with }\;\; a_{-n} = a_n\;\;\forall n\in\mathbb{Z}. $$

Then, for example,

$$ a_3 = 5283 = (-144\cdot 1\cdot 31 + 2257\cdot 3^2)/3. $$

Answer 3

For a = b = c there are infinitely many solutions. See this wolfram alpha page: https://demonstrations.wolfram.com/PythagoreanTrianglesWithTheSameArea/

The parameterization is as follows:

m = r^2 + r*s + s^2

n = r^2 - s^2

u = r^2 + r*s + s^2

v = 2rs + s^2

p = r^2 + 2rs

q = r^2 + r*s + s^2