LTE problem 1

Number Theory Level 4

Find all positive integers $n>1$ such that $2^n \mid 3^n-1.$ Answer as summation of all solutions.

The answer is 6.

1 solution

Kushal Bose
May 30, 2017

The highest exponent of a prime $p$ in factorization of $n$ is $v_p(n)$

For prime $p=2$ the highest exponent is $v_2(x^n-y^n)=v_2(x+y) + v_2(x-y) + v_2(n)-1 \\=v_2(3+1) + v_2(3-1) + v_2(n)-1 \\ =v_2(4) + v_2(2) + v_2(n)-1 \\=2+1+v_2(n)-1=v_2(n)+2$

The highest exponent of $2$ in $3^n-1$ is $v_2(n)+2$

So, if $2^n|3^n-1$ then $v_2(n)+2 \geq n \implies v_2(n) \geq n-2$

The only solutions are $n=2,4$

LTE problem 1

The answer is 6.

1 solution

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