Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
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

1 Answer 1

up vote 8 down vote accepted

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

share|improve this answer
    
@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 at 0:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.