Determine if the Diophantine equation $$x^{2008}-y^{2008}=2^{2009}$$ has any solutions.

What I tried was to look if both sides of the equations would have the same remainder $\pmod{4}$ and use this fact to see if there is any solutions. Since $n^2 \equiv0,1,2 \pmod{4}$ one could write the equation as $$(x^2)^{1004}-(y^2)^{1004}=4\cdot2^{2007}.$$

But this doesn’t seem to help. I’ve also noted that the $LHS$ is just the difference of squares which could be written as $$(x^{1004}-y^{1004})(x^{1004}+y^{1004})$$ but couldn’t find anything to do with this property. What are the alternatives here?

  • 2
    $\begingroup$ Try to solve $a^{2008}-b^{2008}=2.$ $\endgroup$
    – user376343
    Sep 13, 2020 at 18:55

2 Answers 2


The Diophantine equation to check is

$$x^{2008} - y^{2008} = 2^{2009} \tag{1}\label{eq1A}$$

It's clear the parity of $x$ and $y$ must be the same. Consider if they are both even, say $x = 2x'$ and $y = 2y'$. Then \eqref{eq1A} becomes, as user376343's question comment suggests,

$$2^{2008}(x')^{2008} - 2^{2008}(y')^{2008} = 2^{2009} \implies (x')^{2008} - (y')^{2008} = 2 \tag{2}\label{eq2A}$$

However, if $x' = \pm 1$ and $y' = 0$, then you get a result of $1$, while for any other values of $x'$ and $y'$ you will get, e.g., as suggested by the binomial theorem expansion, a difference of much more than $2008$ and, in particular, more than $2$.

This means that $x$ and $y$ must both be odd. Then, as you've shown, the left side of \eqref{eq1A} can be factored to get

$$(x^{1004} - y^{1004})(x^{1004} + y^{1004}) = 2^{2009} \tag{3}\label{eq3A}$$

Note $x^{1004} \equiv y^{1004} \equiv 1 \pmod{4} \implies x^{1004} + y^{1004} \equiv 2 \pmod{4}$. Thus, $x^{1004} + y^{1004}$ has just one factor of $2$. As such, unless $x, y = \pm 1$, which gives a value of $0$ in \eqref{eq1A}, then $x^{1004} + y^{1004}$ has an odd factor greater than $1$. However, the right side of \eqref{eq3A} is a power of $2$, so this is not possible.

In conclusion, there are no integer solutions to \eqref{eq1A}.


Suppose there is a solution in non-negative integers (if there is a solution including a negative integer, say $x=n<0$, we can simply replace $n$ by $|n|>0$ since its exponent is even).

Since the right-hand side of the equation is even and positive, $x$ and $y$ must be of the same parity, with $x > y$. $y$ cannot be zero, otherwise the equation reduces to $x^{2008}=2(2^{2008})$ implying $(x/2)^{2008}=2$ which clearly does not hold for any integer $x$.

Suppose now that $(x,y)=(3,1)$. This does not provide a solution since :

$$x^{2008}-y^{2008}=3^{2008}-1 > 3^32^{2005}-1 > (3^3-1)2^{2005})>2^4(2^{2005})=2^{2009}$$

Generalising, suppose $(x,y)=(m+2,m)$ for some $m \geq 1$. This cannot be a solution since for any exponent $k>1$, the larger an integer $N$ the larger is the gap between $N^k$ and $(N+1)^k$, and similarly for the gap between $N^k$ and $(N+2)^k$, so that:

$$(m+2)^{2008}-m^{2008}\geq 3^{2008}-1 > 2^{2009}$$

Generalising further to include all remaining possibilities, suppose $(x,y)=(m+a,m)$ where $a\geq2$. Then $(m+a)^{2008}\geq(m+2)^{2008}$ and therefore:

$$(m+a)^{2008}-m^{2008}\geq(m+2)^{2008}-m^{2008} > 2^{2009}$$

So the equation has no solutions in integers.


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