Divisibility of $(4^{2^{2n+1}}-3)$ by 13.

Question

The question is to simply proof that $13\space|(\space4^{2^{2n+1}}-3)$ for all $n\in\mathbb{N}$. Since the chapter in my book is about the Euler-phi function and Euler's theorem, my guess would be to look at $4^{2^{2n+1}}\equiv3\space \pmod{13}$ and do something with Euler's theorem. Another guess is to solve this by induction, but in both approaches I am very stuck.

Answer 1

As you correctly note, to compute $$ 4^{2^{2n+1}} \pmod{13} $$ you can compute $$ 2^{2n+1} \pmod{12}. $$ It is advisable to use CRT, and compute separately modulo $3$ and modulo $4$.

Modulo $3$ you have $$ 2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pmod{3}, $$ whereas modulo $4$ you have $$ 2^{2n+1} \equiv 0 \pmod{4}, $$ for $n \ge 1$.

Therefore $$ 2^{2n+1} \equiv 8 \pmod{12}. $$ (This usually implies solving the system of congruences $$ \begin{cases} x \equiv -1 \pmod{3}\\ x \equiv 0 \pmod{4}\\ \end{cases} $$ but in this case it is just a matter of checking which of $x = 0, 4, 8$ is a solution.)

Thus $$ 4^{2^{2n+1}} \equiv 4^8 \equiv 3^4 \equiv 3 \pmod{13} $$ as $4^2 = 16 \equiv 3 \pmod{13}$ and $3^3 = 27 \equiv 1 \pmod{13}$.

Answer 2

${\rm mod}\ 13\!:\,\ 4^{\large 2^{\Large 2n+1}}\!\!\!= 4^{\large 2\cdot 4^{\Large n}}\!\!= 16^{\large 4^{\Large n}}\!\!\equiv 3^{\large 4^{\Large n}}.\ $ But $\ {3^{\large 4}\!\equiv 3}\ $ so $\ \color{#c00}{3^{\large 4^{\Large n}}\!\!\equiv 3}\ $ by induction, viz.

$$ 3^{\large 4^{\Large n+1}}\!\!\equiv (\color{#c00}{3^{\large 4^{\Large n}}})^{\large 4}\equiv \color{#c00}3^{\large 4}\equiv 3$$

Answer 3

First check for $n=0$: $$4^{2^{2\cdot0+1}}=4^2=16 \equiv 3 \pmod{13}$$

Then assume, that $4^{2^{2k+1}} \equiv 3 \pmod{13}$ for $k\geq 0$

$$4^{2^{2(k+1)+1}}= 4^{2^{2k+1+2}}= 4^{2^{2k+1}\cdot 4}=\left(4^{2^{2k+1}}\right)^4 \equiv 3^4 \pmod{13} \equiv (-4)^2 \pmod{13} \equiv 3 \pmod{13}$$

Using the mathematical induction we've proved that the equation $4^{2^{2n+1}}\equiv3\space \pmod{13}$ is true for all $n\geq 0$

Answer 4

Here is another way of viewing the solution.

You are looking at $$4^{2\times 2\times \cdots \times 2}=(4^2)^{2^\cdots}\pmod {13}.$$

You know that $4^2\equiv3\pmod {13}$, $3^2\equiv9\pmod{13}$ and $9^2\equiv (-4)^2\equiv 3\pmod {13}$.

So if the number of twos is odd (which is the case here), then

$$4^{2^{2n+1}}\equiv 3\pmod {13}.$$

Answer 5

By the binomial theorem, $$ 4^{2^{2n+1}}=4^{2\cdot 2^{2n}}=16^{2^{2n}}=(13+3)^{2^{2n}}=13a+3^{2^{2n}} $$ Now $3^{3} \equiv 1 \bmod 13$ and so we have to compute $2^{2n} \bmod 3$. But this is easy: $$ 2^{2n} = 4^n \equiv 1^n =1 \bmod 3 $$ Therefore, $$ 4^{2^{2n+1}} \equiv 3 \bmod 13 $$

Answer 6

As you are aware from looking at Euler's theorem, exponents follow a cycle of values in modular arithmetic. Euler's theorem tells you that $4$ follows a cycle of at most $\phi(13)=12$ under $\bmod 13$ arthimetic, but you can immediately see that $4$ is a square and so will follow a cycle of at most $6$ (because its root(s) must follow a cycle of $12$).

So we need to find the value of $2^{2n+1} \bmod 6$ to find out where we are in this cycle. Since $4^n\equiv 4 \bmod 6$ for any $n\in \Bbb N$, we can see that likewise $2^{2n+1} \equiv 2 \bmod 6$ for any $n$.

So $4^{2^{2n+1}}\equiv 4^2\equiv 3 \bmod 13$ as required.

Answer 7

You can prove this by induction.

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First, show that this is true for $n=0$:

$4^{2^{2\cdot0+1}}=13+3$

Second, assume that this is true for $n$:

$4^{2^{2n+1}}=13k+3$

Third, prove that this is true for $n+1$:

$4^{2^{2(n+1)+1}}=$

$4^{2^{2n+2+1}}=$

$4^{2^{2n+1+2}}=$

$4^{2^{2n+1}\cdot2^2}=$

$4^{2^{2n+1}\cdot4}=$

$(\color\red{4^{2^{2n+1}}})^4=$

$(\color\red{13k+3})^{4}=$

$28561k^4+26364k^3+9126k^2+1404k+81=$

$28561k^4+26364k^3+9126k^2+1404k+78+3=$

$13(2197k^4+2028k^3+702k^2+108k+6)+3$

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Please note that the assumption is used only in the part marked red.