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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...

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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.

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    $\begingroup$ Peter, Bravo! Thanks for explaining the clue to your technique. $\endgroup$ – Hamid Reza Ebrahimi Jul 8 '16 at 9:20
  • $\begingroup$ The tip below the break is epic; I was just about to post a comment on "how to tackle this type of modularity problems in general." $\endgroup$ – Frenzy Li Jul 8 '16 at 9:22
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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 ?

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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.

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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.

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    $\begingroup$ This approach points naturally towards Peter's mod 8 observation: if $x$ and $y$ have opposite parity, then clearly LHS is odd. But if $x$ and $y$ have the same parity then the LHS is the product of three even numbers (in fact it is divisible by $2^4$). This has very little to do with with $499$ being prime and instead is about $499$ being odd. $\endgroup$ – Erick Wong Jul 8 '16 at 13:14
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    $\begingroup$ The algebraic factorization demands that the number be factorizable as product of $3$ numbers $abc$ with $b-a$ even. A number of the form $4p$ does not admit such a factorization because p is prime. If it were merely odd, say, product of $17$ and $23$ my argument breaks down. So my argument needs primality of 499. $\endgroup$ – P Vanchinathan Jul 8 '16 at 13:43

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