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I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it? This my proof: $$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$ Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$ Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.

Domenico Vuono
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  • It suffices to observe that since all the exponents are odd, $n^7-n^3 \equiv 0 \bmod 3$, and $n^{21}-n^{13} \equiv 0 \bmod 3$. – Brian Tung May 01 '15 at 19:02
  • The powers are so high that using induction and binomial theorem becomes quite tedious. I'm sure it's possible, but it won't be short. – Arthur May 01 '15 at 19:02
  • @Arthur I agree. Since the powers are so high, but the expression is modulo a very small number, doing observation in modulo arithmetic is reasonable. – MonkeyKing May 01 '15 at 19:04

5 Answers5

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Another approach

we see that $n(n-1)(n+1)$ divides both $n^7-n^3=n^3(n-1)(n+1)(n^2+1)$ and $n^{21}-n^{13}=n^{13}(n-1)(n+1)(n^2+1)(n^4+1)$

and we know that $(n-1)n(n+1)$ is a multiple of $3$ because the product of three consecutive integers is divisible by $3$.

Elaqqad
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Clear if $\,\color{#a0f}{n\equiv 0},\ $ else $\!\!\!\! \overbrace{\begin{align}\color{#0a0}{n^2}\equiv(\pm1)^2\equiv \color{#0a0}1\\ n^{\color{#c00}{1+2k}}\!\equiv n(\color{#0a0}{n^2})^k\!\equiv\color{#c00} n\end{align}}^{\Large\qquad\quad n\ \ \equiv\ \ \pm 1\ \ {\rm if}\ \ \color{#a0f}{n\, \not \equiv\, 0}}\!\! $ so $\,\ n^{\large \color{#c00}7}-n^{\large \color{#c00}3}\equiv\, \overbrace{\color{#c00}n-\color{#c00}n}^{\large 0}\,\equiv\, n^{\large \color{#c00}{21}}-n^{\large \color{#c00}{13}}$

Alternatively verify $\,f(-1)\equiv f(0)\equiv f(1)\equiv 0\,$ so $\,f(n)\equiv 0,\,$ by $\,n\equiv 0,\pm1\pmod 3.\,$ Some other answers essentially do that by showing $\,(n-1)n(n+1)\mid f(n).\,$ Both are equivalent by the Factor Theorem. Therefore we deduce

Theorem $ $ If $\,f(x)\,$ is a polynomial with integer coefficients then $\,3\mid f(n)\,$ for all integers $\,n\,$ $\iff $ its constant term $\,f(0),\,$ coef sum $\,f(1),\,$ alternating coef sum $\,f(-1)\,$ are all divisible by $3$

Compare to the Parity Root Test.

Community
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Bill Dubuque
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$$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}=n^6(n^2-1)(n^2+1)^2+n^{13}(n^8-1)=$$

$$=n^6\left((n^2-1)(n^2+1)^2+n^7(n^2-1)(n^2+1)(n^4+1)\right)=$$

$$=n^6(n^2-1)(n^2+1)\left[n^2+1+n^7(n^4+1)\right]=$$

$$=\color{red}{n(n-1)(n+1)}\cdot n^5(n^2+1)\left[n^2+1+n^7(n^4+1)\right]$$

Can you see why the red part is already a multiple of $\;3\;$ ?

Timbuc
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The given proof is incomplete, as Fermat's Little Theorem tells that for $p$ prime $n^{p-1} \equiv 1 \pmod{p}$ for $n$ and $p$ co-prime. Of course the proof given can be fixed, as one always has a factor of $n$.

What is always true though is that $n^{p} \equiv n \pmod{p}$

Thus, the result can be proved by noting that $n^3 -n$ divides the expression. This is the same polynomial as $(n-1)n(n+1)$, so that one can see that the argument is not much different from using Fermat Little theorem.

quid
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I will prove the induction method here for $n \geq 0$

Base case : $n=0,1$

Clearly $0$ works.

now $(1^7-1^3)(1^5 + 1^3) + 1^{21} - 1^13 = 0$ which is also a multiple of $3$

Now assume it works for a value $k \geq 0$ that is $$(k^7-k^3)(k^5 + k^3) + k^{21} - k^{13}$$ is a mulitple of $3$

That means that $3m = (k^7-k^3)(k^5 + k^3) + k^{21} - k^{13} = k^{21} -k^{13} + k^{12} + k^{10} -k^{8} -k^{6} $ for some integer $m$

Now we just need to show that this holds for $(k+1)$. That is we need to show that $(k+1)^{21} -(k+1)^{13} + (k+1)^{12} + (k+1)^{10} -(k+1)^{8} -(k+1)^{6}$ is a multiple of $3$

Can you continue from here ??

alkabary
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