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I have proved this in following way.

Assume that $ 2^n \equiv 1 \pmod{n} $.

that means $n\mid(2^n -1)$.

but by proof by contradiction, for $n=3$ this does not hold and we can say $n \nmid (2^n-1) $.

Hence our earlier assumption is wrong.

So, by contradiction we can say that $2^n \not \equiv 1 \pmod{n}$ for any $n$.

My question is, Is this the correct formal proof ? Can I consider this as correct formal proof?

EDIT: If this is not correct, please provide one.

dustin
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CodeLover
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    No. The negation of the statement is that there exists some $n$ such that $2^n \equiv 1 \mod n$. Showing that this is false for the case $n = 3$ doesn't contradict this. – Simon S Nov 22 '14 at 02:53
  • using n=3 i just dis proved the fact that $ n|2^n -1 $. Not the whole proof. – CodeLover Nov 22 '14 at 02:55
  • No this doesn't prove what you want. In fact, you just showed that the statement holds for n=3. – August Nov 22 '14 at 02:56
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    This is an insidious and subtle logical error. Again: what you want to show is that for all positive integer $n$, $2^n \not\equiv 1 \mod n$. The negation of that statement is: there exists some $n$ such that $2^n \equiv 1 \mod n$. Showing that this last statement is false for the case $n = 3$ does not show there is no such $n$ at all; only that it is not the case for $n = 3$. – Simon S Nov 22 '14 at 02:57
  • Understood. Please provide a correct solution. – CodeLover Nov 22 '14 at 03:01
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    "So, by contradiction we can say that $2^n\not\equiv1\pmod n$ for some $n$." – peterwhy Nov 22 '14 at 03:04
  • Note for $n=1$ the initial statement does hold for 1 dividing anything. – JB King Nov 22 '14 at 03:12
  • That's my mistake. Thanks corrected. – CodeLover Nov 22 '14 at 03:13
  • yes. It is avail. – CodeLover Nov 22 '14 at 03:29
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    "For any" is an English term that is not well defined in mathematics in the presence of a negation. Your sentence can be read as "2^n \equiv 1 \mod{n} is false for all $n \gt 1$" or as "2^n \equiv 1 \mod{n} is false for some $n \gt 1$. If the statement were positive it would clearly be "for all $n \gt 1$ – Ross Millikan Nov 22 '14 at 03:56
  • I would also add that "proof by contradiction" and "disproof by counterexample" are different things. – Robert Soupe Nov 22 '14 at 05:32

3 Answers3

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We have $2^1\equiv 1\pmod{1}$. Not very interesting! We show there are no others.

For suppose the congruence $2^n\equiv 1\pmod{n}$ holds, and $n\gt 1$. Let $p$ be the smallest prime divisor of $n$.

Then by Fermat's theorem, $$2^{p-1}\equiv 1\pmod{p}.$$ Also, since $p$ divides $n$, we have $$2^n\equiv 1\pmod{p}.$$

Let $d=\gcd(n,p-1)$. If $d\gt 1$ then $n$ has a divisor greater than $1$ but less than $p$, contradicting the choice of $p$.

Thus $d=1$, and therefore there exist integers $x$ and $y$ such that $(p-1)x+ny=1$. Since $2^{p-1} \equiv 1\pmod{p}$ and $2^{n}\equiv 1\pmod p$, we conclude that $2^1=2^{(p-1)x+ny}\equiv 1 \pmod{p}$. But $2^1\equiv 1\pmod{p}$ is impossible.

André Nicolas
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  • Nice! This approach uses the fact that"if $gcd(n,p-1) = 1$, there exist integers x and y such that $(p−1)x+ny=1$". Do we have another approach? – SiXUlm Nov 22 '14 at 07:12
  • Yes, the Bezout "identity." I do not know another one, though this one can be reworded to make use of properties of order mod $p$. – André Nicolas Nov 22 '14 at 07:16
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"For any" is an English term that is not well defined in mathematics in the presence of a negation. Your sentence can be read as "$2^n \equiv 1 \mod{n}$ is false for all $n \gt 1$" or as "$2^n \equiv 1 \mod{n}$ is false for some $n \gt 1$. If the statement were positive it would clearly be "for all $n \gt 1$" If you read it as "$2^n \equiv 1 \mod{n}$ is false for some $n \gt 1$ the example you have shown is sufficient. If you read it as "$2^n \equiv 1 \mod{n}$ is false for all $n \gt 1$" one example is not sufficient-you need to prove the failure for all $n \gt 1$

Ross Millikan
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Here's another pass at this:

Suppose for some $k$, we have a positive integer $m$ such that $2^k-1 = mk$. Or, $2^k \equiv 1 (m)$

Therefore, in the vector space $\mathbb{F}_2^k$, there are $mk$ non-zero vectors. Partition these into $m$ disjoint sets of $k$ vectors: $S_i$ for $i=1,2, \ldots m$. Choose vectors $v_i \in S_i$ to form $\{v_1, v_2, \ldots v_k\}$. Now for any other vector, $w$, not in this set, the collection $\{v_1, v_2, \ldots v_k\}\cup\{w\}$ is linearly dependent. Therefore, the set spans $\mathbb{F}_2^k$ and thus it is a basis. Therefore, the number of possible bases is $m^k$. But then, $o(GL_k(\mathbb{F}_2)) = (2^k-1)(2^k-2)\ldots(2^k-2^{k-1})$

Older attempt

Hope this is right. Suppose there is a $k$ such that $2^k - 1 \equiv 0 (k)$.

That is, $1+2+2^2+ \ldots +2^{k-1} \equiv 0 (k)$, or, adding the two up:

$2+2^2+\ldots +2^k \equiv 0 (k)$

But this can't be since $(2,k)=1$ as, by hypothesis, $2$ is invertible in $\mathbb{Z}/k\mathbb{Z}$

past_his_bedtimeman
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