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

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?

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

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.