Actually I have done this problem by induction.(trivial)
Please tell me another method to do this problem instead of induction.
Here are some other divisibility statements and I have also done these by induction. But tell me another method without induction.
a) $21 | 4^{n+1}+5^{2n-1}$
b) $24| 2\cdot7^n+3\cdot 5^n-5$
i wrote correct problem it was wrong edited by Kunnysan
Answer to a). First prove its divisible by $3$. Using binomial expansion, we can show $$ 4^{n + 1} + 5^{2n - 1} \equiv (3 + 1)^{n + 1} + (3 + 2)^{2n - 1} \equiv 1^{n + 1} + 2^{2n - 1} \equiv 1 + (3 - 1)^{2n - 1} \equiv 1 + (-1)^{2n - 1} \equiv 1 - 1 \equiv 0 \mod 3 $$ Now prove that it is divisible by $7$ as follows. $$ 4^{n + 1} + 5^{2n - 1} \equiv 2^{2n + 2} + (7 - 2)^{2n - 1} \equiv 2^{2n + 2} + (-2)^{2n - 1} \equiv 2^{2n + 2} - 2^{2n - 1} \equiv 2^{2n - 1}(2^3 - 1) \\ \equiv 7 \cdot 2^{2n - 1} \equiv 0 \mod 7 $$ So $4^{n + 1} + 5^{2n - 1}$ is divisible by both $3$ and $7$ (which are coprime) which means it is divisible by $21$.
Answer to b). First prove it is divisible by $6$. $$ 2\cdot7^n+3\cdot 5^n-5 \equiv 2 \cdot (6 + 1)^n + 3 \cdot (6 - 1)^n - 5 \equiv 2 \cdot 1^n + 3 \cdot (-1)^n - 5 \equiv 2 - 5 \pm 3 \\ \equiv -3 \pm 3 \equiv 0 \mod 6 $$ Now prove it is divisible by $8$. $$ 2\cdot7^n+3\cdot 5^n-5 \equiv 2 \cdot (8 - 1)^n + 3 \cdot (8 - 3)^n - 5 \equiv 2(-1)^n + 3(-3)^n - 5 \mod 8 $$ When $n$ is even (say $n = 2k$), we have $$ 2 + 3^{2k + 1} - 5 \equiv 3(9^k - 1) \equiv 3(9 - 1)(9^{k - 1} + \dotsb + 1) \equiv 0 \mod 8 $$ When $n$ is odd, $n + 1$ is even (say $n + 1 = 2k$) and so we have $$ -2 - 3^{2k} - 5 \equiv -(9^k + 7) \equiv -((8 + 1)^k + 7) \equiv -(1 + 7) \equiv 0 \mod 8 $$ Divisibility by $6$ implies divisibility by $3$. Since $3$ and $8$ are coprime, we have divisibility by $24$.
Here's a proof that your statement is wrong.
First note this: $3^{2n} = 9^n$
If n is even then $9^n \equiv 1 (\mod 5)$, otherwise it's $9^n \equiv -1 (\mod 5)$
Case 1: n is even
We need to prove that:
$$9^n + 2^n + 1 + 1 \equiv 0 (\mod 5)$$, from the upper relations we have $$2^n \equiv 2 (\mod 5)$$
This one is true only if $n=4k+1$, for $k = 0,1,2,3...$ But this means n is odd which isn't possible because we assume n is even.
Case 2: n is odd
$$9^n + 2^n + 1 + 1 \equiv 0 (\mod 5)$$ $$2^n \equiv 4 (\mod 5)$$
This one is true only for $n=4k+2$ for $k = 0,1,2,3...$ But this is not possible because at the beginning we assum that n is odd.
This means that for any $n = {1,2,3,4,5...}$, this statement is wrong.
Maybe you want a proof that 5 isn't factor of the expression?
Hints:
$2\cdot7^n++3\cdot5^n-5=2(7^n-1)+3(5^n-1)=12(7^{n-1}+...+1)+12(5^{n-1}+...+1)=12(2+(5+7)+...+(5^{n-1}+7^{n-1}))$
The term inside bracket is even always.
$4^{n+1}+5^{2n-1}=16\cdot4^{n-1}+5\cdot25^{n-1}\equiv 5\cdot(-4)^{n-1}-5\cdot4^{n-1}....$ continue.
HINT: $$4^{n+1}+5^{2n-1}=4^2(4^{n-1})+5(5^2)^{n-1}=16\cdot4^{n-1}+5(25)^{n-1}$$
Now using congruence formula#$10$ here, $25\equiv4\pmod{21},25^{n-1}\equiv4^{n-1}\pmod{21}$ (Proof below)
Can you take it home from here?
For $2\cdot7^n+3\cdot 5^n-5$
Observe that $5^2=25\equiv1\pmod {24}$ and $7^2=49\equiv1 \pmod {24}$
So, if $n$ is odd, $5^n\equiv5,7^n\equiv7\pmod{24}$
and for $n$ even, $5^n\equiv1,7^n\equiv1\pmod{24}$
Proofs:
$1:$ We know, $(a-b)|(a^n-b^n)$ Proof (1,2)
So, $25^m-4^m$ will be divisible by $25-4=21$ for integer $m\ge0$
$2:$ $$(4+21)^{n-1}=4^{n-1}+\binom{n-1}14^{n-2}21+\cdots+\binom{n-1}{n-2}4\cdot21^{n-2}+21^{n-1} $$ $$=4^{n-1}+21\cdot\text{ some integer} $$ as binomial coefficients are integer(Proof)
While your main question is getting corrected, we look at another one of the questions in order to supply you with additional tools which will turn out to be useful after the correction. for ultimately the fact that $3\equiv -2\pmod{5}$ will turn out to be handy.
(a) We show that $21$ divides $N$, where $N=4^{n+1}+5^{2n-1}$. It is enough to show that $3$ divides $N$ and $7$ divides $N$.
1. First we deal with divisibility by $3$. Note that $4\equiv 1\pmod{3}$ and $5\equiv -1\pmod{3}$. It follows that $$N\equiv (1)^{n+1}+(-1)^{2n-1}\pmod{3}.$$ But $(1)^{n+1}\equiv 1$, and $(-1)^{2n-1}\equiv -1$, since $2n-1$ is odd. It follows that $N\equiv 0\pmod{3}$.
2. Next we deal with divisibility by $7$. Note that $N=2^{2n+2}+5^{2n-1}=8\cdot 2^{2n-1}+5^{2n-1}$. But $5\equiv -2\pmod{7}$. It follows that $$N=8\cdot 2^{2n-1}+5^{2n-1}\equiv 2^{2n-1}+(-2)^{2n-1}\equiv 0\pmod{7}.$$
(b) The calculation modulo $3$ is immediate. We have $2\cdot7^n+3\cdot 5^n-5\equiv 2-5\equiv 0\pmod{3}$.
For the calculation modulo $8$. it may be best to separate the cases $n$ even and $n$ odd. If $n$ is even, then $2\cdot 7^n\equiv 2$, and $3\cdot 5^n\equiv 3$, and thus our expression is congruent to $2+3-5$.
If $n$ is odd then $2\cdot 7^n\equiv -2$, and $3\cdot 5^n\equiv -1$, and thus our expression is congruent to $(-2)+(-1)+(-5)$.
