Prove that $x^4-y^4=1996$ has no integer root. 
Prove that $x^4-y^4=1996$ has no integer root.

$LHS=(x-y)(x+y)(x^2+y^2)=1996$
Now we have to consider all possible decompositions of $1996$  resulting in a non Linear system of equations seemingly complicated...
 A: Since $1996=2^2\cdot 499$, there are not many ways to write $1996$ as the product of these three factors (and possible trivial factors). For example, try $x-y=1$, $x+y=4$, $x^2+y^2=499$. Here we can use that $499$ cannot be the sum of two squares, because $499\equiv 3\bmod 4$. For more details see this question. Indeed, what divisor of $1996$ can $x^2+y^2$ be ?
A: Using your factorization gives
$(x-y)(x+y)(x^2+y^2)=1996=4\times499$
When $x$ and $y$ are integers the first two terms in  LHS gives two factors differing by an even number.
As 499 is a prime number, it is clear that $4\times499$ cannot be written as product of 3 numbers with 2 of them differing by an even number.
EDIT: The factorization $1\times 2\times 998$ indeed has two factors  differing by an even number.  In this case $x-y$ is $1$ and  $x+y=4$, which is impossible in integers.
A: Here is another approach using less arithmetic and more analysis (inequalities). The only arithmetic that is needed is: $x$ and $y$ should both be even or odd as rhs is 1996.
That is they must differ by at least 2.
First let us see the possibility of $x$ and $y$ differing by $2$.
Call them say $t+2$ and $t$. 
Clearly the function  $(t+2)^4-t^4$ is an increasing function for $t>1$. (Verified easily using derivatives) 
So $(t+2)^4-t^4=1996$ can have just one solution with $t>1$.
As $8^4-6^4=2800$, just check for  for $t=1,2,\ldots, 7$ we cannot get a difference of 1996.
Next try $(t+4)^4-t^4$ we need to check even smaller range for $t$. . . .
This is is a finite verification method, effective, though ugly.
A: Look at it modulo 8.
For any integer $n$, $n^4$ will be either 0 or 1 mod 8:


*

*if $n$ is even, then $n$ is divisible by 2, so $n^4$ is divisible by $2^4$, so $n^4 = 0$ mod 8

*if $n$ is odd, then it’s either 1, 3, 5, or 7 mod 8; it’s easy to check that each of these yields 1 when raised to the 4th power.


So then for any integers $x$ and $y$, $x^4 - y^4$ can only be 0, 1, or –1 (mod 8).  But 1996 is 4 (mod 8).

How I found this solution: for problems like this, modular arithmetic can often provide a simple “obstruction” in this way, so it’s always a good direction to try.  So I looked at the possibilities for $x^4, y^4$ under several small moduli — 2, 4, and then 8 — and found that 8 gave a sufficiently restrictive set of possibilities, as hoped.
