In a 1995 paper, Choudhry gave a table of solutions to the quartic Diophantine equation,

$a^4+nb^4 = c^4+nd^4\tag{1}$

for $n\leq101$. Seiji Tomita recently extended this to $n<1000$ and solved all $n$ except $n=967$ (which was later found by Andrew Bremner).

It can be shown there is an infinite number of $n$ such that $(1)$ is solvable, such as $n=v^2-3$. In general, given the Chebyshev polynomials of the first kind $T_m(x)$

$$\begin{aligned} T_1(x) &= x\\ T_2(x) &= 2x^2-1\\ T_3(x) &= 4x^3-3x\\ \end{aligned}$$

etc, and the second kind $U_m(x)$

$$\begin{aligned} U_1(x) &= 2x\\ U_2(x) &= 4x^2-1\\ U_3(x) &= 8x^3-4x\\ \end{aligned}$$

etc, define, $$n = \frac{T_m(x)}{x}$$ $$y = \frac{U_{m-2}(x)}{x}$$

then one can observe that,

$(n + n x + y)^4 + n(1 - n x - x y)^4 = (n - n x + y)^4 + n(1 + n x + x y)^4\tag{2}$

** Questions**:

- Is $(1)$ solvable in the integers for
positive integer $n$ where $a^4\neq c^4$? (This has also been asked in this MSE post.)*all* - How do we show that $(2)$ is indeed true for all positive integer $m$?