Difference between revisions of "1964 IMO Problems/Problem 1"

Revision as of 23:04, 16 August 2013

(a) Find all positive integers $n$ for which $2^n-1$ is divisible by $7$ .

(b) Prove that there is no positive integer $n$ for which $2^n+1$ is divisible by $7$ .

We see that $2^n$ is equivalent to $2, 4,$ and $1$ $\pmod{7}$ for $n$ congruent to $1$ , $2$ , and $0$ $\pmod{3}$ , respectively.

(a) From the statement above, only $n$ divisible by $3$ work.

(b) Again from the statement above, $2^n$ can never be congruent to $-1$ $\pmod{7}$ , so there are no solutions for $n$ .

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 == Solution ==
 We see that <math>2^n</math> is equivalent to <math>2, 4,</math> and <math>1</math> <math>\pmod{7}</math> for <math>n</math> congruent to <math>1</math>, <math>2</math>, and <math>0</math> <math>\pmod{3}</math>, respectively.
-(a) From the statement above, only <math>n</math> divisible by <math>3</math> work.
+'''(a)''' From the statement above, only <math>n</math> divisible by <math>3</math> work.
-(b) Again from the statement above, <math>2^n</math> can never be congruent to <math>-1</math> <math>\pmod{7}</math>, so there are no solutions for <math>n</math>.
+'''(b)''' Again from the statement above, <math>2^n</math> can never be congruent to <math>-1</math> <math>\pmod{7}</math>, so there are no solutions for <math>n</math>.