LTE applications
$begingroup$
Let k be a positive integer. Find all positive integers n such that
$3^k | 2^n - 1$
I just started reading and learning about the Lifting the Exponents Lemma, and I want to try and use it in this problem.
So I know that if that said statement is true, then
$V_3(2^n - 1) > 0$
I guess you can express it as$V_3(2^n - 1^n)$, but in the lifting the exponent lemmas, I can't find one theorem that can solve the above expression, since 3 does not divide 2-1.
diophantine-equations exponentiation
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
$endgroup$
add a comment |
$begingroup$
Let k be a positive integer. Find all positive integers n such that
$3^k | 2^n - 1$
I just started reading and learning about the Lifting the Exponents Lemma, and I want to try and use it in this problem.
So I know that if that said statement is true, then
$V_3(2^n - 1) > 0$
I guess you can express it as$V_3(2^n - 1^n)$, but in the lifting the exponent lemmas, I can't find one theorem that can solve the above expression, since 3 does not divide 2-1.
diophantine-equations exponentiation
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
$endgroup$
1
$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48
add a comment |
$begingroup$
Let k be a positive integer. Find all positive integers n such that
$3^k | 2^n - 1$
I just started reading and learning about the Lifting the Exponents Lemma, and I want to try and use it in this problem.
So I know that if that said statement is true, then
$V_3(2^n - 1) > 0$
I guess you can express it as$V_3(2^n - 1^n)$, but in the lifting the exponent lemmas, I can't find one theorem that can solve the above expression, since 3 does not divide 2-1.
diophantine-equations exponentiation
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
$endgroup$
Let k be a positive integer. Find all positive integers n such that
$3^k | 2^n - 1$
I just started reading and learning about the Lifting the Exponents Lemma, and I want to try and use it in this problem.
So I know that if that said statement is true, then
$V_3(2^n - 1) > 0$
I guess you can express it as$V_3(2^n - 1^n)$, but in the lifting the exponent lemmas, I can't find one theorem that can solve the above expression, since 3 does not divide 2-1.
diophantine-equations exponentiation
diophantine-equations exponentiation
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
asked Jan 21 at 13:25
SuperMage1SuperMage1
897211
897211
1
$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48
add a comment |
1
$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48
1
1
$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48
$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48
add a comment |
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$begingroup$
Right, but $3|2^2-1$.
$endgroup$
– Mindlack
Jan 21 at 13:48