MHB Remainder of $3^{2^n}-1$ Divided by $2^{n+3}$

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Remainder
AI Thread Summary
The remainder of \(3^{2^n} - 1\) when divided by \(2^{n+3}\) is \(2^{n+2}\). This conclusion is reached through mathematical induction, starting with the base case where \(n=1\), confirming that \(3^{2^1} - 1 = 8\) is an odd multiple of \(2^3\). The inductive step shows that \(3^{2^{n+1}} - 1\) can be expressed as the product of two factors, where the first is an odd multiple of \(2^{n+2}\) and the second is also an odd multiple of \(2\). The discussion highlights the clarity of the proof provided by a user named Opalg, which is appreciated by others in the thread. Overall, the mathematical reasoning is solid and effectively demonstrates the claim.
kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
find the remainder when $3^{2^n}-1$ is divided by $2^{n+3}$
 
Mathematics news on Phys.org
kaliprasad said:
find the remainder when $3^{2^n}-1$ is divided by $2^{n+3}$
[sp]Claim: $3^{2^n}-1$ is an odd multiple of $2^{n+2}$.

Proof by induction: Base case ($n=1$) is true because $3^{2^1}-1 = 3^2-1 = 8$, which is an odd multiple of $2^{1+2} = 2^3 = 8.$ For the inductive step, $3^{2^{n+1}} - 1 = (3^{2^n}-1) (3^{2^n}+1)$ (difference of two squares). By the inductive hypothesis, the first factor is an odd multiple of $2^{n+2}$. The second factor is $(3^{2^n}-1) + 2$ which, again by the inductive hypothesis, is an odd multiple of $2$. Therefore the product $3^{2^{n+1}} - 1$ of the two factors is an odd multiple of $2^{n+3}$, which completes the inductive step.

It follows that the remainder when $3^{2^n}-1$ is divided by $2^{n+3}$ is $2^{n+2}$.[/sp]
 
It's been over a decade since I've done these kinds of problems on a regular basis - and I can't top Opalg's delicious proof above (nicely done! (Sun) ) - but I'm just wondering, would it not be easier if you were to...

write $$3^{2^n}-1$$ as $$9^n-1$$ and $$2^{n+3}$$ as $$8\cdot 2^n$$

?
 
DreamWeaver said:
It's been over a decade since I've done these kinds of problems on a regular basis - and I can't top Opalg's delicious proof above (nicely done! (Sun) ) - but I'm just wondering, would it not be easier if you were to...

write $$3^{2^n}-1$$ as $$9^n-1$$ and $$2^{n+3}$$ as $$8\cdot 2^n$$

?

[sp]$$3^{2^n} = 3^{(2^n)} \ne (3^2)^n = 3^{2n} = 9^n$$[/sp]
 
Bacterius said:
[sp]$$3^{2^n} = 3^{(2^n)} \ne (3^2)^n = 3^{2n} = 9^n$$[/sp]

Egg on face! Ha ha! Thanks for that... I don't know what I was thinking there... :o:o:o
 
good solution by opalg
here is another
$3^{2^n} - 1 = (3^{2^{n-1}} +1) (3^{2^{n-1}} -1)$
= $(3^{2^{n-1}} +1) (3^{2^{n-2}} +1)(3^{2^{n-2}} -1$
= $(3^{2^{n-1}} +1) (3^{2^{n-2}} +1)\cdots(3^2+1)(3^2-1)$

now there are 1st n-1 terms of the form $3^{2^k} + 1 $ where k = n-1 to 1

$3^{2^ k} - 1 = 3^{2x} - 1 = 9^x - 1$ as $2^k$ is even so

$3^{2^k} -1 \mod 4 = 0$

$3^{2^k} +1 \mod 4 = 2$

so $3^{2^k} +1 = 4 m + 2 = 2(2m+ 1)$

there are n-1 numbers of the form 2(2m+1) and last number is 8.
so product = $8 * 2^{n-1} * (2y + 1) = y . 2^{n+3} + 2 ^{n+2}$

so $product \mod 2^{n+3} = 2 ^{n+2}$

hence $2^{n+2}$ is the desired remainder
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top