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Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,

$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$

for $\color{blue}{\text{both}}$ $k = 2,4$ then,

$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k = \\(a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k \tag{2}$$

for $k = 1,3,5,7$.

The system $(1)$ be equivalently expressed as,

$$\begin{align} x_1^k+x_2^k+x_3^k\, &= y_1^k+y_2^k+y_3^k,\quad \color{blue}{\text{both}}\; k = 2,4\\ x_1+x_2-x_3\, &= 2(y_1+y_2-y_3)\\ x_1+x_2-x_3\, &\ne 0\tag{3} \end{align}$$

There are only two quadratic parameterizations known so far to $(3)$, namely,

$$(-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = \\(9x-y)^k + (-x+3y)^k + (16x-2y)^k\tag{4}$$

where $126x^2-5y^2 = z^2$ and,

$$(6x+3y)^k + (4x+9y)^k + (2x-12y)^k = \\(-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k\tag{5}$$

where $x^2+10y^2 = z^2$ found by Sinha and (yours truly). The square-free discriminants are $D = 70, -10$, respectively.

Question: Any other solution for $(3)$ in terms of quadratic forms?

P.S. There are a whole bunch of elliptic curves that can solve $(3)$.

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  • $\begingroup$ I trust you've looked at Gloden's book, and there's nothing there? $\endgroup$ – Gerry Myerson Nov 25 '14 at 6:19
  • $\begingroup$ I read Sinha's paper (which came out decades after Gloden's book). Sinha makes no mention of Gloden, so I assume the result is his. $\endgroup$ – Tito Piezas III Nov 25 '14 at 7:11
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As regards the system of equations $(3).$

The system of equations. $$\left\{\begin{aligned}& a^2+b^2+c^2=x^2+y^2+z^2\\&a+b-c=2(x+y-z)\end{aligned}\right.$$

Solutions have the form: $$a=4t(p+k-s)+6p^2+2k^2+8kp-6ps-2ks$$ $$b=t^2+2t(p+k-s)+3p^2-3k^2-6s^2-2kp-6ps+8ks$$ $$c=t^2+2t(p+k-s)-3p^2+3k^2-6s^2-2kp+2ks$$ $$x=2t(p+k-s)+6p^2-2k^2-6s^2+4kp+8ks$$ $$y=t^2+2t(p+k-s)+3p^2+3k^2+4kp-12ps-4ks$$ $$z=t^2+2t(p+k-s)+3p^2+3k^2-6s^2+4kp-6ps+2ks$$

$t,p,k,s$ - integers asked us.

For the system of equations: $$\left\{\begin{aligned}&a^3+q^3+c^3=n^3+k^3+r^3\\&a+q-c=2(n+k-r)\end{aligned}\right.$$

Solutions have the form. $$a=6(2x-2b+3y-z)(b^2+yz-yb-zb)$$ $$***$$

$$q=14b^3+z^3-7x^3-27y^3+12(z+2b-3y)x^2-$$ $$-6(4b^2+9y^2+z^2-12by+4bz-6yz)x+$$ $$+3zb^2-45yb^2+57by^2+9bz^2-24yzb+24zy^2-12yz^2$$
$$***$$

$$c=2b^3+z^3-7x^3-27y^3+12(z+2b-3y)x^2-$$ $$-6(4b^2+9y^2+z^2-12by+4bz-6yz)x+$$ $$+21zb^2-27yb^2+51by^2+3bz^2-48ybz+30zy^2-6yz^2$$

$$***$$

$$n=6x(b^2+yz-yb-zb)$$

$$***$$

$$k=8b^3+z^3-7x^3-27y^3+12(z+2b-3y)x^2-$$ $$-6(4b^2+9y^2+z^2-12by+4bz-6yz)x+$$ $$+9zb^2-33yb^2+51by^2+9bz^2-36yzb+30zy^2-12yz^2$$

$$***$$

$$r=8b^3+z^3-7x^3-27y^3+12(z+2b-3y)x^2-$$ $$-6(4b^2+9y^2+z^2-12by+4bz-6yz)x+$$ $$+15zb^2-39yb^2+57by^2+3bz^2-36yzb+24zy^2-6yz^2$$

$b,z,x,y$ - integers asked us.

You can write a solution and more. But I don't see the point. It is too bulky. When I find easy - I will write.

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  • $\begingroup$ Why do you solve $a^3+q^3+c^3=n^3+k^3+r^3$, when OP didn't ask about that equation? OP is interested in $a^4+b^4+c^4=d^4+e^4+f^4$. $\endgroup$ – Gerry Myerson Nov 25 '14 at 6:18
  • $\begingroup$ Individ, it's the language factor again. You need to solve $a^k+b^k+c^k = d^k+e^k+f^k$ true for both $k = 2,4$. To make it crystal clear, the system has THREE eqns, $$a^2+b^2+c^2 = d^2+e^2+f^2$$ $$a^4+b^4+c^4 = d^4+e^4+f^4$$ $$a+b-c = 2(d+e-f)$$ $\endgroup$ – Tito Piezas III Nov 25 '14 at 7:05
  • $\begingroup$ @TitoPiezasIII understand which system of equations. $\endgroup$ – individ Nov 25 '14 at 9:17
  • $\begingroup$ @Individ: Sigh, is this a question? Those three eqns CLEARLY above your comment is the system that needs to be solved. Sigh. $\endgroup$ – Tito Piezas III Nov 25 '14 at 12:41
  • $\begingroup$ @TitoPiezasIII The solution can be written, but it is bulky. Even Wolfram alpha can't write a formula completely. I will try to write it compactly. $\endgroup$ – individ Nov 25 '14 at 12:44
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I think not to introduce additional equations, and directly solve the system of equations.

$$\left\{\begin{aligned}&R^2+Q^2+T^2=X^2+Y^2+Z^2\\&R^4+Q^4+T^4=X^4+Y^4+Z^4\end{aligned}\right.$$

Using integer parameters $k,s,t$ - Will make a replacement.

$$a=3(k+s-t)^2+k(k-t)$$

$$b=3(k+s-t)^2+s(s-t)$$

$$c=3(k+s-t)^2-t^2+(k+s)t-2ks$$

$$x=3(k+s-t)^2-ks$$

$$y=3(k+s-t)^2-t^2+(k+s)t-ks$$

$$z=3(k+s-t)^2+k^2+s^2-(k+s)t$$

Then the solution can be written as:

$$R=3a^4+(4a-b)b^3+(4a-c)c^3-(4a-x)x^3-(4a-y)y^3-(4a-z)z^3$$

$$Q=(4b-a)a^3+3b^4+(4b-c)c^3-(4b-x)x^3-(4b-y)y^3-(4b-z)z^3$$

$$T=(4c-a)a^3+(4c-b)b^3+3c^4-(4c-x)x^3-(4c-y)y^3-(4c-z)z^3$$

$$X=(4x-a)a^3+(4x-b)b^3+(4x-c)c^3-3x^4-(4x-y)y^3-(4x-z)z^3$$

$$Y=(4y-a)a^3+(4y-b)b^3+(4y-c)c^3-(4y-x)x^3-3y^4-(4y-z)z^3$$

$$Z=(4z-a)a^3+(4z-b)b^3+(4z-c)c^3-(4z-x)x^3-(4z-y)y^3-3z^4$$

To obtain relatively Prime solutions - after substitution should be reduced to common divisor.

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  • $\begingroup$ @TitoPiezasIII Let the formula be? maybe then someone will come in handy. $\endgroup$ – individ Nov 26 '14 at 16:56
  • $\begingroup$ Sorry, Individ, but I just found that this also satisfies, $$R+Q+T = X+Y+Z = 0$$ which, since $(3)$ involves even powers, is equivalent to, $$R+Q-(-T) = X+Y-(-Z) = 0$$ and prohibited by Sinha's theorem. :( $\endgroup$ – Tito Piezas III Nov 27 '14 at 1:21
  • $\begingroup$ @TitoPiezasIII I therefore this equation and replaced it. This condition does not allow to solve the equation. $\endgroup$ – individ Nov 27 '14 at 4:37
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(Too long for a comment.)

I simplified your expression and found they are ternary quadratic forms. (Why didn't you just simplify them? Maple and Mathematica can do it easily.) So,

$$R^n+Q^n+T^n = X^n+Y^n+Z^n,\quad for\; n =2,4\tag{1}$$

$$\begin{align}R =& -2 k^2 - 2 k s + s^2 + 3 k t - t^2\\ Q =&\; k^2 - 2 k s - 2 s^2 + 3 s t - t^2\\ T =&\; k^2 + 4 k s + s^2 - 3 k t - 3 s t + 2 t^2\\ X =&\; k^2 + k s + s^2 - t^2\\ Y =&\; k^2 + k s + s^2 - 3 k t - 3 s t + 2 t^2\\ Z =& -2 k^2 - 2 k s - 2 s^2 + 3 k t + 3 s t - t^2 \end{align}$$

Unfortunately, this also satisfies,

$$R+Q+T =X+Y+Z = 0$$

or equivalently, since $(1)$ involves even powers,

$$R+Q-(-T) =X+Y-(-Z) = 0$$

a case prohibited by Sinha's theorem.

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