Prove by Induction $19^{n}-4^{2n}$ is divisible by 3

Question

Assume $n=0$ for $19^{n}-4^{2n}$

Just going to do some simplification first: $19^{n}-4^{2n}\implies19^{n}-16^{n}$ and if $n=0$ then $19^{0}-16^{0}$ then it is zero which is obviously divisible by 3.

Now for the inductive hypothesis $19^{n}-16^{n}\mod{3}=0$

So now I show it for the case $n+1$ and one gets: $19^{n+1}-16^{n+1}\implies19^{n}\cdot19-16^{n}\cdot16$ but from here how do I show divisibility by 3?

Make the appropriate modifications to the answer over here, or over here. Also see the questions linked to the first of these. — Sarvesh Ravichandran Iyer, May 24 '23 at 13:42
Try this: $a^{n+1}-b^{n+1}=(a+b)(a^n-b^n)-ab(a^{n-1}-b^{n-1})$, and from case $n-1$, $n$ prove the case $n+1$ — Jz Pan, May 24 '23 at 13:44
With $f(n)=19^{n}-4^{2n}$. we get $f(n+1)-f(n)=3 \cdot ( 6 \cdot 19^n-5 \cdot 4^{2n})$ — lhf, May 24 '23 at 16:35

score 3 · Accepted Answer · answered May 24 '23 at 13:44

3

Let $19^n-16^n=3k$ then $$19^{n+1}-16^{n+1}$$ $$=19^n\cdot19-16^n\cdot16$$ $$=(3k+16^n)19-16^n\cdot16$$ $$=57k+19\cdot16^n-16\cdot16^n$$ $$=57k+3\cdot16^n$$ $$=3\lambda$$

answered May 24 '23 at 13:44

MathStackexchangeIsNotSoBad

4,716

Third step is a bit confusing: $19^{n}\cdot19\implies(3k+16^{n})19$ – lifeiscool May 24 '23 at 13:46
2

I assumed that $19^n-16^n=3k$ then adding $16^n$ to both sides of this equation yeilds $19^n=3k+16^n$ and I substituted this value in the third step – MathStackexchangeIsNotSoBad May 24 '23 at 13:51
2

OOOHHHH ok, I got it now, thanks :) – lifeiscool May 24 '23 at 13:53

score 3 · Answer 2 · answered May 24 '23 at 14:23

So you’re off to a good start. The next step is just to realize that $19-16=3$, which is the reason this is true at all.

So you have: $$19(19^n) - 16(16^n) = (16+3)19^n - 16(16^n) = 16(19^n - 16^n)+3(19^n)$$

Now from your induction assumption $(19^n - 16^n)$ is a multiple of 3, and $3(19^n)$ is clearly a multiple of 3. You’re adding two multiples of 3, the result is a multiple of 3

Prove by Induction $19^{n}-4^{2n}$ is divisible by 3

2 Answers2