1964 IMO Problems/Problem 1


(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 $7$.

Solution 1

We claim $2^n$ 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 $3$ will work.

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

Solution 1.1

This solution is clearer and easier to understand.

(1) Since we know that $2^n-1$ is congruent to 0 (mod 7), we know that $2^n$ is congruent to 8 mod 7, which means $2^n$ is congruent to 1 mod 7.

Experimenting with the residue of $2^n$ mod 7:

$n$=1: 2

$n$=2: 4

$n$=3: 1 (this is because when $2^n$ is doubled to $2*2^n$, the residue doubles too, but $4*2=8$ is congruent to 1 (mod 7).

$n$=4: 2

$n$=5: 4

$n$=6: 1

Through induction, we easily show that this is true since the residue doubles every time you double $2^n$.

So, the residue of $2^n$ mod 7 cycles in 2, 4, 1. Therefore, $n$ must be a multiple of 3. Proved.

(2) According to part (1), the residue of $2^n$ cycles in 2, 4, 1.

If $2^n+1$ is congruent to 0 mod 7, then $2^n$ must be congruent to 6 mod 7, but this is not possible due to how $2^n$ mod 7 cycles. Therefore, there is no solution. Proved.


See Also

1964 IMO (Problems) • Resources
Preceded by
First question
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions