How to solve this diophantine equation ($4$ variables)? $$4(9k^4-l^4)=9m^4-n^4$$
Well, it seems a little bit harder than finding Pythagorean triples. Right now I have no idea where to actually start so any help on this would be appreciated.
 A: We show by descent that $k=l=m=n=0$ is the only solution.  
Modulo $5$, the left-hand side is congruent to $k^4+l^4$, while the right-hand side is congruent to $-(m^4+n^4)$. 
Any fourth power is congruent to $0$ or $1$ modulo $5$. So the sum of two fourth powers is congruent to $0$, $1$, or $2$. 
In particular, the sum of two fourth powers is congruent to the negative of a sum of two fourth powers only if both sums are divisible by $5$. And this only happens if $k$, $l$, $m$, and $n$ are all divisible by $5$. 
So if $(k,l,m,n)$ is a solution, then so is $(k/5,l/5,m/5,n/5)$. Continue. We conclude that $k$, $l$, $m$, and $n$ are divisible by arbitrarily high powers of $5$, so are all $0$.
Remark: The classical Frmat-style descent argument can be reworded. Suppose that there is a solution $(k,l,m,n)$ with not all the variables equal to $0$. Then there is a solution with $|k|+|l|+|m|+|n|$ positive and as small as possible. Use the above argument to conclude that $(k/5,l/5,m/5,n/5)$ is an integer solution. This contradicts the minimality of $|k|+|l|+|m|+|n|$. Or else we can reword things so as to use standard mathematical induction terminology.  
