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?

• What is it that makes them "harmonic"? The fact that $c=d^2-1$? – Arturo Magidin Jul 7 '12 at 19:26
• @Arturo Magidin, We set a=y^2, b=(y+1)^2, c=(y^2+y+2)y(y+1), d=(y^2+y+1)^2. – Vassilis Parassidis Jul 7 '12 at 19:43
• @VassilisParassidis I think you should the general form of the harmonic solution to your question. I thought you were asking if there were infinite solutions to the original equation. – user17762 Jul 7 '12 at 19:47
• A quick search using gp/pari 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. – Old John Jul 7 '12 at 19:59

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 $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$

• @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 $7^4=6^4+3^4+32^2$. – Vassilis Parassidis Feb 9 '14 at 0:22