1. ## Mathematical Induction

8 divides $5^n-4n-1$ for all integers n greater than or equal to 1.

Proof:
Basis step:
n = 1

$5^1-4(1)-1$ = 0
0/8 = 0
8 divides $5^n-4n-1$ if n=1

Induction step:
Assume $5^n-4n-1$ is divisible by 8 (induction hypothesis)
[show 5^(n+1) - 4(n+1) - 1 is divisible by 8]

• $5^(n+1) - 4(n+1) - 1 = 5^n+1 - 4n + 4 - 1$
• $= 5^n+1 - 4n +3$
• $=5^n *5^1 - 4n +3$

...now im not sure where to go

2. ...sorry the first bullet should be 5^n+1 - 4(n+1) -1 = ....

3. damn, and the second one should be 5^n+1 - 4n + 3

4. A few points.

1. You can use the "Edit" feature rather than posting multiple times, it will keep the thread cleaner.

2. You made a silly error of not distributing the -4 properly, so you should have -5 where you currently have +3.

3. I think the problem can be solved by considering that when n is odd, 4n+1 is congruent to 5 (mod 8), and when n is even, 4n+1 is congruent to 1 (mod 8). That is, we have by induction hypothesis

$5^n-(4n+1)\equiv0\ (\text{mod}\ 8)$

and so we can find out what $5^n$ is (mod 8), by considering the two cases, n odd or even.

4. The way to get exponents with more than one character in LaTeX is like this (hover mouse over it to see the code): $x^{12345}$.

Edit: I had a typo; the -4 in red above used to say -1.

5. Originally Posted by luckyNUM7
8 divides $5^n-4n-1$ for all integers n greater than or equal to 1.

Proof:
Basis step:
n = 1

$5^1-4(1)-1$ = 0
0/8 = 0
8 divides $5^n-4n-1$ if n=1

Induction step:
Assume $5^n-4n-1$ is divisible by 8 (induction hypothesis)
[show 5^(n+1) - 4(n+1) - 1 is divisible by 8]
Since $5^n-4n-1$ is divisible by 8, then $5^n-4n-1=8x, x\in \mathbb{Z}$

Multiply through by 5

$5\cdot 5^n-5\cdot 4n-5\cdot 1=5\cdot 8x$

$5^{n+1}-4n-16n-4-1=5\cdot 8x$

Move $-16n$ to RHS

$5^{n+1}-(4n-4)-1=16n+5\cdot 8x$

$5^{n+1}-4(n+1)-1=8(2n+5x)$

Since $2n+5x$is an integer, $5^{n+1}-4(n+1)-1$ is divisible by 8.