Equation $a^{n}+b^{n}=2008$ has no integers solutions.

Question

Prove that the equation $a^{n}+b^{n}=2008$ has no solutions for $a,b,n\in\mathbb{Z}, n\geq2.$

Answer 1

The bad and brute-force approach would be like this:

You see that $2^{11} > 2008$, so you only have to consider $ 2 \leq n \leq 10$ It will become tedious, but you will probably finish it in about 30-50 min.

You can neglect $n = 6$ pretty soon from FLT, since $2008 \equiv 6 \bmod 7$ and $a^6 + b^6 \equiv (0,1) + (0,1) \not\equiv 6 \bmod 7$

Same goes for $n = 4$ and $n = 10$, checking with $\bmod 5$ and $\bmod 11$ respectively.

Answer 2

Not an answer but a simplification perhaps:

Well $$a^n+b^n\equiv 2008\equiv 0 \text{ mod 2}$$ $$\implies a^n\equiv -b^n\equiv b^n \text{ mod 2}$$ $$\implies a\equiv b \text{ mod 2}$$

Thus either both $a$ and $b$ are both odd or they are both even.

Now suppose both $a$ and $b$ are even then we can write:

$$a=2c \text{ and } b=2d \text{ so that we get:}$$

$$(2c)^n+(2d)^n=2008$$

$$\implies 2^n(c^n+d^n)=2008$$ $$\implies 2^n\mid 2008$$

But sense: $2008=2^3\times 251$ we know that $n$ can not exceed $3$.

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Thus we can deduce from this that either both $a$ and $b$ are even or they are both odd. If they are both even then it suffices to show that there are no integer solutions $c$ and $d$ to:

$$c^2+d^2=502 \text{ and } c^3+d^3=251$$

Answer 3

Case 1 $n=2k$ is even. Then, $a^{2k},b^{2k} \equiv 0,1,4 \pmod{8}$ and therefore both $a,b$ need to be even.

Then $a^{2k}, b^{2k}$ are both divisible by $2^{2k}$. Thus $2^{2k}|2008$ which implies that $k=1$.

Writing $a=2a', b=2b'$ we have

$$a'^2+b'^2=502 \,.$$

Taking this equation modulo 8, there are no solutions.

Case 2 $n=2k+1$ is odd. Then

$$(a+b)(a^{2k}-a^{2k-1}b+...+b^{2k})=2008 \,.$$

If $a,b$ have opposite parity, then both $a+b$ and $a^{2k}-a^{2k-1}b+...+b^{2k}$ are odd, which is not possible.

Therefore, $a,b$ are both odd or both even.

Subcase 2a $a,b$ are both even. Then $2^{2k+1}|a^{2k+1}+b^{2k+1}=2008$ which implies that $2k+1=3$.

Writing $a=2a', b=2b'$ we have

$$a'^3+b'^3=251 \,.$$

Then $$(a'+b')(a'^2-a'b'+b'^2)=251 \,.$$

Using the fact that $251$ is prime, there are only two possible factorizations, and in each case there is no solution.

Subcase 2b $a,b$ are both odd. Then $a+b$ is even and $(a^{2k}-a^{2k-1}b+...+b^{2k})$ is odd. As 251 is prime, there are only two possibilities

$$a+b=8$$ $$a^{2k}-a^{2k-1}b+...+b^{2k}=251$$ or $$a+b=2008$$ $$a^{2k}-a^{2k-1}b+...+b^{2k}=1$$

At least one of $a$ and $b$ has to be positive.

• If both $a,b$ are positive, it is easy to show that $a+b=2008, a^{2k+1}+b^{2k+1}=2008$ is impossible.

The equation $a+b=8, a^{2k+1}+b^{2k+1}=2008$ leads to only two possibilities: $$7^{2k+1}+1=2008$$ or $$3^{2k+1}+5^{2k+1}=2008$$ which can easily be eliminated.

• If only one of $a,b$ is positive, by symmetry, we can assume that $a >0, b<0$. Let $c=-b >0$.

Then the two equations reduce to $$(8+c)^{2k+1}-c^{2k+1}=2008$$ or $$(2008+c)^{2k+1}-c^{2k+1}=2008$$ The second one can easely be eliminated by observing that $$(2008+c)^{2k+1} > 2008^{2k+1}+c^{2k+1}$$

For the first, $$2008=(8+c)^{2k+1}-c^{2k+1} >8^{2k+1}+c^{2k+1}-c^{2k+1}=8^{2k+1}$$

This implies that $k=1$ and it is easy to show that $$(8+c)^3-c^3=2008$$ has no solution with $c>0$.