4

Problem from Burton's Number Theory:

Let $n>1$ be an odd integer . Show that $n\nmid 3^n+1.$

Trying by induction on $n$. The result holds for $n=3 $ since $3\nmid 3^3+1=28$.

Let the result hold for $n=m\implies m \nmid 3^m+1.$

Assume contrary, let $m+1\mid 3^{m+1}+1\implies 3^{m+1}+1=k(m+1)\implies 3.3^m=km+(k-1)$.

I am unable to proceed further.Please help.

Learnmore
  • 31,675
  • 10
  • 110
  • 250

2 Answers2

5

Proof by Contradiction. $$n|3^n+1 \implies n| (n-3)^n-1$$ Now let the smallest prime factor of $n$ be $p$. Note that by Fermat's Little Theorem $$p|(n-3)^{p-1}-1$$ We also have $$p|(n-3)^n-1$$ So we have that $$\begin{align}p|\gcd((n-3)^{p-1}-1, (n-3)^n-1) & \\ \implies (n-3)^{\gcd(p-1,n)}-1=n-4 \equiv 0 \pmod {p} \end{align}$$

This follows from here and the fact that $p$ is the smallest prime divisor.

So we have $p|4$. But $p$ is odd. Contradiction.

Community
  • 1
S.C.B.
  • 23,106
  • You mean $3^n+1 \equiv 0 \pmod {n} \implies (n-3)^n-1 \equiv 0 \pmod {n}$? It follows from $(n-3)^n-1 \equiv -3^n-1 \equiv 0 \pmod{n}$. – S.C.B. Jan 20 '17 at 03:52
  • We can eliminate use of $,\gcd(a^{\large J}!-1,a^{\large K}!-1) = a^{\large\gcd(J,K)}!-1,$ by instead examining the order of $,3,$ mod $,p,,$ e.g. see my answer. In fact one can prove this gcd result the same way. – Bill Dubuque Jan 20 '17 at 05:13
  • @BillDubuque I thought using orders would be overkill. I learned about this proof far earlier, before knowing about orders. It is simple enough without that. I already know about the proof you linked. I just think this is a better method. – S.C.B. Jan 20 '17 at 07:45
  • It is worth becoming proficient with order-based proofs as early as possible since they are much more general.They are ubiquitous in number theory and algebra - something I emphasize in my answers by often highlighting the key role played by order and denominator ideals.This innate algebraic structure is clarified when one studies abstract algebra. – Bill Dubuque Jan 20 '17 at 15:19
  • @BillDubuque OK, you know best, I guess. I am just a 15 year old student after all, and I don't believe I am going to major in maths. – S.C.B. Jan 20 '17 at 15:38
  • Your understanding of number theory is quite good for that age. If you enjoy it then your should consider learning more advanced number theory and algebra since there lies much beauty. By the way, I added a remark to my answer which explains in more detail how the proofs are related. Best of luck in your studies. – Bill Dubuque Jan 20 '17 at 16:04
4

Let $\,\color{#0a0}{p= \rm least}$ prime factor of $n.\,$ ${\rm mod}\ p\!:\ 3^{\large n}\!\equiv -1\,\Rightarrow\,3^{\large 2n}\!\equiv 1\equiv 3^{\large p-1}$ so the order $\,k\,$ of $3$ satisfies $\,k\mid 2n,p\!-\!1$ so $\,\color{#c00}k\mid(2n,p\!-\!1)=(2,p\!-\!1)=\color{#c00}2\,$ by $\,n\,$ is coprime to $p\!-\!1\,$ (by $\,n\,$ has no prime $\color{#0a0}{{\rm factors}\ < p}).\,$ Thus $\,3^{\large\color{#c00}2}\!\equiv 1\pmod{\!p}\,$ so $\,p\mid 3^{\large 2}\!-1=8,\,$ contra $\,p\,$ odd, by $\,p\mid n$ odd.

Remark $\ $ The proof in S.C.B's answer is closely related. We can eliminate the $2$ in $2n$ as follows. Note $\, 1 \equiv -3^{\large n}\!\equiv (-3)^{\large n}\!\equiv (-3)^{\large p-1}$ so $\,-3\,$ has order $\,\color{#c00}{k\!=\!1},\,$ by $\,k\,$ divides the coprimes $\,n,\,p\!-\!1.\,$ Thus $\,(-3)^{\large\color{#c00}{\bf 1}}\!\equiv 1\,\Rightarrow\, 4\equiv 0\,\Rightarrow\,p\mid 4,\,$ contra $\,p\,$ odd.

Finally replace $\,-3\,$ by a positive rep $\,n\!-\!3\equiv -3\pmod{\!p}\,$ then replace the above logic on orders mod $\,p\,$ by equivalent gcd logic in the integers using $ \,\gcd(a^{\large J}\!-1,a^{\large K}\!-1) = a^{\large\gcd(J,K)}\!-1.$

Community
  • 1
Bill Dubuque
  • 282,220