How do I prove that 5*7^n + 3*11^n -8 is divisible by 3 for all natural numbers n using congruence modulo properties?

To prove :

[math]3\text{ | } (5\cdot 7^n +3 \cdot 11^n -8)[/math]

i.e., To prove :

[math](5\cdot 7^n +3\cdot 11^n -8) \equiv 0 \pmod 3[/math]

Proof :

We know, [math]7 \equiv 1 \pmod 3[/math]

Raising both sides to power [math]n,[/math]

[math] 7^n \equiv 1 \pmod 3[/math]

Multiplying both sides by [math]5,[/math]

[math] 5\cdot 7^n \equiv 5 \pmod 3 \quad ...(i)[/math]

Again, [math]11 \equiv 2 \pmod 3[/math]

Raising both sides to power [math]n,[/math]

[math]11^n \equiv 2^n \pmod 3[/math]

Multiplying both sides by [math]3,[/math]

[math]3\cdot 11^n \equiv 3\cdot 2^n \pmod 3 \equiv 0 \pmod 3 \quad ...(ii)[/math]

[math]\Big[\because[/math] [math]0[/math] is the remainder obtained on dividing [math](3 \cdot 2^n)[/math] by [math]3[/math] i.e., [math]3 \text{ | } (3\cdot 2^n)[/math] [math].\Big][/math]

Also, [math]8 \equiv 5 \pmod 3 \quad ...(iii)[/math]

Now, adding congruence [math](i)[/math] & [math](ii)[/math] and subtracting congruence [math](iii),[/math] we get,

[math](5\cdot 7^n +3\cdot 11^n -8) \equiv (5+0-5) \pmod 3[/math]

[math]\boxed{\therefore (5\cdot 7^n +3\cdot 11^n -8) \equiv 0 \pmod 3 .} \quad \dagger[/math]