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From my work on hyperelliptic equations I found how to get infinitely many solutions of the equation $a^4+b^4+c^2=d^4$. I call these solutions harmonic: $\begin{array}{rcccccl} 1^4 &+& 2^4 &+& 8^2 &=& 3^4\\ 2^4 &+& 3^4 &+& 48^2 &=& 7^4\\ 3^4 &+& 4^4 &+& 168^2 &=& 13^4\\ 4^4 &+& 5^4 &+& 440^2 &=& 21^4 \end{array}$ and so on. All numbers natural.

Does anyone know if there are infinite non harmonic solutions of this equation?

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    A quick se$a$rch using gp/p$a$ri seems to show lots of solutions not in your list of "harmonic" ones, so I would expect there to be an infinite number of non-harmonic solutions.2012-07-07

1 Answers 1

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EDIT

There seems to be a lot of non-harmonic solutions.

Here are couple of one parameter family of non-harmonic solutions.


$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2 = (a^2 - a + 1)^2$ You could scale these up appropriately i.e. $\left(ka,ka(a-1),k^2(a-1)(2a^2 - a + 1),k \left(a^2-a+1 \right) \right),$ to get other solutions.


$b = a(a+1), d = a^2 + a + 1$ and $c = (a+1)(2a^2+a+1)$ which relies on the identity $a^4 + \left( a(a+1)\right)^4 + \left((a+1)(2a^2+a+1) \right)^2 = (a^2 + a + 1)^2$ You could again scale these up appropriately i.e. $\left(ka,ka(a+1),k^2(a+1)(2a^2 + a + 1),k \left(a^2+a+1 \right) \right),$ to get other solutions.


You could also take your harmonic solution $(a,a+1,a(a+1)(a^2 + a + 2),a^2+a+1)$ and scale appropriately, i.e. $\left(ka,k(a+1),k^2a(a+1)(a^2 + a + 2),k \left(a^2+a+1 \right) \right),$ to get other solutions.


Yes. Below is a one parameter family of infinite solutions.

$b = a+1$, $d = a^2+a+1$, $c = (a^2+a+1)^2-1$.

$b^4 = (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$

$d^4 = (a^2 + a +1)^4 = 1+4 a+10 a^2+16 a^3+19 a^4+16 a^5+10 a^6+4 a^7+a^8$

Hence, \begin{align} d^4 - b^4 - a^4 & = 4 a^2+12 a^3+17 a^4+16 a^5+10 a^6+4 a^7+a^8\\ & = a^2 \left(4 +12 a+17 a^2+16 a^3+10 a^4+4 a^5+a^6 \right)\\ & = a^2 (a+1)^2 (a^2 + a + 2)^2 \end{align} Hence, choose $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$

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    @user17762.From the formulas you propose we obtain two nontrivial solutions of the equation $d^4=b^4+c^4+f^2$ for every value of $a=2$ or a>2, when all numbers positive integers. eg. We obtain two solutions when $a=2$: $7^4=3^4+2^4+48^2$, $7^4=6^4+2^4+33^2$, but we do not get the solution $7^4=6^4+3^4+32^2$.2014-02-09