Prove by induction that for all natural numbers $n$, $\frac{5}{4}8^n + 3^{3n-1}$ is divisible by $19$.

I'm running into trouble at the inductive part of the step, I am currently attempting to add/subtract the inductive hypothesis but I end up with two different coefficients that are seemingly unrelated to $19$. I've been stuck on this for days, thanks for the help!

  • $\begingroup$ Hint: $8^3\equiv-1\mod19$ and $27^3\equiv-1\mod19$. $\endgroup$
    – Lucian
    Sep 7 '14 at 20:39

Rewrite as $f(n)=10\cdot 8^{n-1}+3^{3n-1}$ to clear the fraction.

Then can you see that $f(n+1)=10\cdot 8^n+3^{3n+2}=8f(n)+(27-8)3^{3n-1}$

We choose the factor $8$ (we could also have chosen $27$) since we want to get rid of one of the exponentials altogether in the $f(n)$ term, and to find that the other one ends up with a coefficient divisible by $19$.

Note that if $f(n)=Aa^n+Bb^n$ then $f(n+1)=(a+b)f(n)-abf(n-1)$ so if $a,b$ are integers and some integer $c$ is a factor of both $f(n-1)$ and $f(n)$ it will also be a factor of $f(n+1)$, which is another way of solving this kind of problem (requires two base cases rather than one, but then the induction goes through).


For $n+1$ step, we have $$\begin{align}\frac 54\cdot 8^{n+1}+3^{3(n+1)-1}&=\frac 54\cdot 8^{n+1}+8\cdot 3^{3n-1} -8\cdot 3^{3n-1}+3^{3n+2}\\&=8\left(\frac 54\cdot 8^n+3^{3n-1}\right)+(3^3-8)\cdot 3^{3n-1}\\&=8\left(\frac 54\cdot 8^n+3^{3n-1}\right)+19\cdot 3^{3n-1}.\end{align}$$

  • $\begingroup$ Ah I see! I was previously trying to add/subtract the entire hypothesis rather than just the second half. Could you maybe explain your thought process that led you to add/subtract 8*3^{3n-1} ? Once I've seen it then it makes sense of course, I'm just thinking about what first led you to that. Thank you again! $\endgroup$
    – Jake
    Sep 7 '14 at 20:42
  • $\begingroup$ @Jake: I wanted the form of $f(n+1)=8f(n)+19\cdot g(n)$ for some $g(n)$. And $8f(n)=\frac 54\cdot 8^{n+1}+8\cdot 3^{3n-1}$. Here, $\frac 54\cdot 8^{n+1}$ is needed for $f(n+1)$, but $8\cdot 3^{3n-1}$ is an 'extra' term, so... make sense? $\endgroup$
    – mathlove
    Sep 7 '14 at 20:57

$\begin{eqnarray}{\bf Hint}\ \ \ {\rm mod}\ 19\!:\,\ \ 3^3\equiv\, 8\ \Rightarrow\ 3^{3n}\!\!&\equiv& 8^n\\ \\ {\rm and}\quad\ 4\,\equiv\, -5\cdot 3\ \Rightarrow\ \dfrac{1}3&\equiv&\!\! {-}\!\!\dfrac{5}4\\ \\ {\rm multiplying \ \, yields}\,\ \ 3^{3n-1}\!&\equiv&\!\! -\!\dfrac{5}4 8^{n}\quad {\bf QED}\end{eqnarray}$

Remark $\ $ As explained here, the proofs in the other answers can be discovered from the above in the following mechanical manner: take the standard proof of the Congruence Product Rule, then substitute the specific numbers in this problem, i.e. simply repeat the proof for the special case at hand. Though many such inductive proofs appear at first glance to be pulled out of a hat like magic, in fact they are, at heart, instances of congruence arithmetic - exactly as above. After one has learned congruence arithmetic the problem can be easily transformed into the triviality $\ (27/8)^n\equiv\, 1^n\equiv 1\ $ since $\ 27\equiv 8\pmod{19 }.$

See also this answer on fractions in modular (congruence) arithmetic.


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