<s>[INST] On $a^4+nb^4 = c^4+nd^4$ and Chebyshev polynomials 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 all positive integer $n$ where $a^4\neq c^4$? (This has also been asked in this MSE post.)

*How do we show that $(2)$ is indeed true for all positive integer $m$?

 [/INST] Question 2:
The reason must be that Chebyshev polynomials solve Fermat-Pell equations.
The difference between the two sides of equation (2),
$$
(n+nx+y)^4 + n(1−nx−xy)^4 = (n−nx+y)^4 + n(1+nx+xy)^4,
$$
factors as $8nx(n+y) \phantom. ((x^2-1)y^2 + 2n(x^2-1)y + 1 - n^2)$.
The last factor is a quadratic in $n$ with leading coefficient $1$,
so has solutions iff the discriminant is a square.
This yields the equation $(x^2-1) u^2 + 1 = t^2$ where $u=xy$,
which is a Fermat-Pell equation in $u$ and $t$
with fundamental solution $(u,t)=(1,x)$, etc.
</s>