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For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that: $$x^n + (x+1)^n = (x+2)^n$$

Writing it using Newton's binom, we obtain: $$x^n = \sum_{i=1}^{n} \binom{n}{i} \cdot x^{n-i} \cdot (2^i - 1)$$

I don't know how to continue the problem. Can you help me, please? Thanks!

gareth618
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    Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case... – Parcly Taxel Jul 18 '18 at 17:42
  • It must be $2^n≡1 \pmod {x}$. – Takahiro Waki Jul 18 '18 at 17:45
  • $$\begin{align}(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^{n-1}+(x+2)^{x-2}x+...+x^{n-1}]\&=2[(x+2)^{n-1}+(x+2)^{x-2}x+...+x^{n-1}]\end{align}$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^{n-1}+(x+2)^{x-2}x+...+x^{n-1}]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$. –  Jul 18 '18 at 17:56

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Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.

Chappers
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  • For $x=3k+1$, we have $x^n \equiv (3k+1)^n\equiv 1\pmod{3}$, $(x+1)^n \equiv (3k+2)^n\equiv 2\pmod{3}$, and $(x+2)^n\equiv(3k+3)^n\equiv 0\equiv 1+2 \pmod{3}$ for odd $n$. – Math Lover Jul 18 '18 at 17:59
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Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.

Claim 2: If $x$ is odd, then $x \equiv \pm 1 \pmod{4}$. In which case (with $n$ being odd) we have left side is either $1+2^n \pmod{4}$ or $-1 \pmod{4}$, whereas the right side is $-1 \pmod{4}$ or $1 \pmod{4}$ respectively. For $n>2$ the two sides are not equal. Hence no solution.

Anurag A
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The reminders by $x$ is equal, then 　

$2^n=kx+1$

$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$

$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$

Since $n≠2n+k$, there doesn't such integers.

Takahiro Waki
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  • This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$. – Batominovski Jul 29 '18 at 12:28