# The equation $a^{4n}+b^{4n}+c^{4n}=2d^2$

Recently, I found that if $$a+b=c$$, then $$a^4+b^4+c^4=2d^2$$ for some positive integer $$d$$. The parametric equation is: $$m^4+n^4+(m+n)^4=2(m^2+mn+n^2)^2$$ The condition $$a+b=c$$ (assuming $$c \geqslant a,b$$) isn't necessary. For example: $$7^4+7^4+12^4=2 \cdot 113^2$$ We can note that when we make the equation in the form $$a^{4n}+b^{4n}+c^{4n}=2d^2$$, and we impose the condition $$a^n+b^n=c^n$$ for the parametric solution:

(i) When $$n=1$$, we can have any positive integers $$a+b=c$$

(ii) When $$n=2$$, we can have any Pythagorean Triple $$(a,b,c)$$.

(iii) When $$n>2$$, there are no solutions by Fermat's Last Theorem.

Checking when $$n=2$$, I saw that there are no solutions for $$a \leqslant b \leqslant c \leqslant 3000$$ where $$a^2+b^2 \neq c^2$$. I have not run a program for any value $$n>2$$ though.

For positive integers $$a \leqslant b \leqslant c$$ where $$\gcd(a,b,c)=1$$ :

$$1$$. Are there any solutions for $$a^8+b^8+c^8=2d^2$$ where $$a^2+b^2 \neq c^2$$ ?

$$2$$. Are there any solutions for $$a^{4n}+b^{4n}+c^{4n}=2d^2$$ where $$n>2$$?

$$3$$. For the solutions of $$a^4+b^4+c^4=2d^2$$ which do not follow $$a+b=c$$, is there any way of generating more solutions from primitive solutions? From primitive solution $$(a,b,c,d)$$, can we get more solutions $$(A,B,C,D)$$?

EDIT : First off, it suffices to focus on solutions for $$a^{4n}+b^{4n}+c^{4n}=2d^2$$ for prime $$n$$ alone, since if we have a solution for some $$n$$, then we have a solution for the divisors of $$n$$ as well. An accepted answer would be one of:

$$(i)$$ Verifying problem $$1$$ for $$a \leqslant b \leqslant c \leqslant 1000000$$.

$$(ii)$$ Verifying problem $$2$$ for $$a \leqslant b \leqslant c \leqslant 100000$$ (for odd primes $$n<100$$).

$$(iii)$$ Verifying problem $$1$$ for $$a \leqslant b \leqslant c \leqslant 100000$$ and problem $$2$$ for $$a \leqslant b \leqslant c \leqslant 10000$$ (for odd primes $$n<100$$).

$$(iv)$$ Proof or Counterexample for either problems $$1$$ or $$2$$.

$$(v)$$ Relations, generation or parametric characterization of the non-trivial solutions of $$a^4+b^4+c^4=2d^2$$

• The equation $$a^8+b^8+c^8=2d^2$$ is derived from: $$\big\{p(x^2-y^2)^4+q\big\}^2+\big\{p(2xy)^4+q\big\}^2+\big\{p(x^2+y^2)^4+q\big\}^2=2\big\{p(x^8+14x^4y^4+y^8)+q\big\}^2+q^2$$ where $(p,q)=(1,0)$ so I don't believe there are alternative solutions to your first question that don't satisfy the constraint. – Mr Pie Jan 21 at 12:27
• It appears that for $n>2$ there is no solution. With which search limit would you be content ? – Peter Jan 21 at 18:29
• For $n>2$, I guess good evidence would be no solution for $a \leqslant b \leqslant c \leqslant 100000$ for $n \leqslant 100$. I wouldn't mind $c \leqslant 10000$ either, but the former is preferable as an answer. For $n=2$, I would be satisfied with no non-Pythagorean solutions for $a \leqslant b \leqslant c \leqslant 1000000$. But I guess that would be almost impossible, so $c \leqslant 100000$ would be okay, but again, the former is preferable as an answer. – Haran Jan 22 at 11:46
• Applying standard equal sums of squares parameterizations (cf. Bradley et al.), there exist integers $m,j,r,s,t$ such that, in the general case, \begin{align} d &= \frac{(mr-js)(r^2+s^2+t^2)}{2(r+s)}, \\ a^{2n} &= t(js-mr), \\ b^{2n} &= \frac{(mr-js)\bigl(t^2-\bigl((r+s)^2-2s^2\bigr)\bigr)}{2(r+s)}, \\ c^{2n} &= \frac{(mr-js)\bigl(t^2+\bigl((r-s)^2-2s^2\bigr)\bigr)}{2(r+s)}, \end{align} or some other permutation of $a,b,c$ (though there may be an easy 'WLOG' argument?). Perhaps you can use this, with your constraint $\gcd(a,b,c)=1$, to move forward? – Kieren MacMillan Jan 22 at 15:55
• select(r->r%9==0 || r%9==2 || r%9==5 || r%9==8,%) %3 = [[0, 0, [0, 0]], [2, 0, [1, 1]], [2, 1, [0, 1]], [5, 0, [1, 4]], [5, 1, [0, 4]], [5, 4, [0, 1]], [8, 0, [1, 7]], [8, 0, [4, 4]], [8, 1, [0, 7]], [8, 4, [0, 4]], [8, 7, [0, 1]], [9, 1, [1, 7]], [9, 1, [4, 4]], [9, 4, [1, 4]], [9, 7, [1, 1]], [11, 0, [4, 7]], [11, 4, [0, 7]], [11, 7, [0, 4]], [14, 0, [7, 7]], [14, 7, [0, 7]], [18, 4, [7, 7]], [18, 7, [4, 7]]] – user645636 Jan 28 at 16:08

Problem 3

This is a scheme to generate the solutions which, like your example of $$(7,7,12,113)$$, have two of $$a,b,c$$ equal.

Consider the following system of three closely related equations.

E: $$2x^4-y^4=z^2$$

F: $$x^4+8y^4=z^2$$

G: $$x^4-2y^4=z^2$$

A 'base solution' $$(x,y,z)$$ of E can be used to generate a solution $$(z,xy,2x^4+y^4)$$ of F.

Each solution $$(x,y,z)$$ of F can be used to generate a solution $$(z,2xy,|x^4-8y^4|)$$ of G.

Each solution $$(x,y,z)$$ of G can be used to generate a further solution $$(z,xy,x^4+2y^4)$$ of F.

Each solution $$(x,y,z)$$ of F can be used to generate the solution $$(x,x,2y,z)$$ of the required equation.

Example starting with the solution $$(1,1,1)$$ of E.

The scheme generates F$$(1,1,3)$$, G$$(3,2,7)$$,F$$(7,6,113)$$, G$$(113,84,7967)$$, F$$(7967,9492,262621633)$$, .....

The required solutions are then $$(1,1,2,3),(7,7,12,113),(7967,7967,18984,262621633),...$$