Is the quartic diophantine equation $a^4+nb^4 = c^4+nd^4$ solvable for any integer $n$? In 1995, Choudhry wrote the paper "On the Diophantine equation $A^4+nB^4 = C^4+nD^4$" and solved it for $75$ $n$ in the range $1\leq n\leq 101$. I thought the missing $n$ were either unsolvable or would have large solutions, but decided to have a crack at it anyway with a simple Mathematica search. 
To my surprise, ALL $n$ in the range had solutions and were of relatively small size. See the complete table here.
So here's the question (also asked by others prior): 

"Is $a^4+nb^4 = c^4+nd^4$ solvable in non-trivial integers (that is $a^4\neq c^4$) for any integer $n$?"


Edit 1 (a day later)
I just learned that Seiji Tomita has a table for all $n<1000$. Only $n=\color{blue}{967}$ is missing.
Edit 2 (a week later)
Andrew Bremner finally found a "small" solution as,
$$416776^4-252386^4=\color{blue}{967}(98431^4-90427^4)$$ 
thus completing the table for $n<1000$. (It is not known if this is the smallest soln and Elkies has searched the range $a,c< 100000$).
Edit 3 (four years later)
It turns out there is a smaller one found by Oleg567,
$$251477^4 - 146927^4 =  \color{blue}{967}(44086^4 - 18748^4)$$
See his answer below for details on $n<5000$.
 A: I used a Maple program to first find the set $S$ of all positive integers $\le 10^{12}$ that can be written as the difference of two fourth powers (there are $649913$ of them),
then for successive integers $n$ find (if it exists) the least member of $S \cap (S/n$).  Thus for $n=2$ the result was $179727600$, corresponding to 
$179727600 = 116^4 - 34^4 = (139^4 - 61^4)/2$.
Hmm, this sequence belongs in the OEIS (it doesn't seem to be there yet).
There were $133$ cases up to $n=1000$ where $S \cap (nS)$ was empty, of which the first was $n=206$.  That doesn't say there is no solution for $n=206$, just that any solution will have $c^4 - a^4 > 10^{12}$.
A: Equation $a^4+nb^4=c^4+nd^4$ has numerical solutions for all $n< 20000$   and is available on      Seiji Tomita website "Computational Number Theory" given below:
http://www.maroon.dti.ne.jp/fermat/eindex.html
A: Above equation given below:
$a^4+nb^4 = c^4+nd^4\tag{1}$
It has been shown that equation $(1)$ has numerical solutions for integer $n$ less than $20000$ and also $n$ rational for $n=\frac{b}a$ , for $100<a<200$ and $1<b<200$ ,
example for $(b,a) =(163,1)$ one solution is $(237,8,89,66)$. These solutions are given in latest article #254, on Seiji Tomita website  www.maroon.dti.ne.jp  If that table is extended $b>200$ than that it will fortify the conjecture that equation $(1)$ above could be true for $n$ both as an integer and a rational number.
