1
$\begingroup$

$$P(n):2304\mid7^{2n}-48n-1$$ I've done the base case; $P(1)$ is true because the expression then evaluates to zero, which is divisible by 2304. Now I'm stuck on the inductive step: proving $P(m+1)$ true if $P(m)$ is true. I do know this though: $$7^{2m+2}-48(m+1)-1 =49\cdot7^{2m}-48m-49$$

$\endgroup$
  • $\begingroup$ Now what have you tried, eh? $\endgroup$ – Parcly Taxel Nov 4 '16 at 2:21
  • $\begingroup$ @ParclyTaxel it's how other sums are solved in my textbook.. please help if you can! $\endgroup$ – user385779 Nov 4 '16 at 2:21
2
$\begingroup$

$$7^{2(m+1)} - 48(m+1) - 1 = 49 \cdot 7^{2m} - 48m - 49 = 49\left(7^{2m} - 48m - 1\right) + 2304m$$

$\endgroup$
  • $\begingroup$ But how can we remove common when it's only 48m and how does 2304 balance it out? Please explain really confused Sir. $\endgroup$ – user385779 Nov 4 '16 at 2:54
  • $\begingroup$ You should verify the equalities I listed, namely that $2304 = 49\cdot 48 - 48$. Then you can continue your induction proof. $\endgroup$ – Michael Biro Nov 4 '16 at 2:58
  • $\begingroup$ THANK YOU SO VERY MUCH! THANKS A LOT SIR. $\endgroup$ – user385779 Nov 4 '16 at 3:02
1
$\begingroup$

Hint $\ $ Conceptually, the induction is simply the first two terms of the Binomial Theorem, namely put $\ a = 48\ $ below and note $\,48^2 = 2304$

${\rm mod}\ a^2\!:\ (1\!+\!a)^n \equiv 1\! +\! na\ $ is true for $\ n = 0\ $ and the induction step is easy:

$\qquad\quad\ (1\!+\!a)^{n+1}\! = (1\!+\!a)(1\!+\!a)^n\overset{\rm induct}= (1\!+\!a)(1\!+\!na) \equiv 1\!+\!(n\!+\!1)a\ $ by $\ a^2\equiv 0$

$\endgroup$
  • $\begingroup$ Wow thanks so explaining so clearly, much appreciated $\endgroup$ – user385779 Nov 4 '16 at 3:51
0
$\begingroup$

Let $f(n)=7^{2n}-48n-1$.

By direct evaluation, we have $f(1)=0$.

We want to show $2304{\,\mid\,}f(n)$, for all positive integers $n$.

Proceed by induction on $n$ . . .

Since $f(1)=0$, the truth for the base case, $n=1$, is immediate.

Next, suppose $2304{\,\mid\,}f(n)$, for some positive integer $n$. \begin{align*} \text{Then}\;\;&2304{\,\mid\,}f(n)\\[4pt] \implies\;&f(n)\equiv 0\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2n}-48n-1\equiv 0\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2n}\equiv 48n+1\;(\text{mod}\;2304)\\[4pt] \implies\;&49\left(7^{2n}\right)\equiv 49(48n+1)\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2(n+1)}\equiv 2352n+49\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2(n+1)}\equiv 48n+49\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2(n+1)}-48n-49\equiv 0\;(\text{mod}\;2304)\\[4pt] \implies\;&7^{2(n+1)}-48(n+1)-1\equiv 0\;(\text{mod}\;2304)\\[4pt] \implies\;&f(n+1)\equiv 0\;(\text{mod}\;2304)\\[4pt] \implies\;&2304{\,\mid\,}f(n+1)\\[4pt] \end{align*} which completes the induction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.