Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$

Question

Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$. I tried solving but got stuck. Show that it is true for n=1 $$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$

Assume it true for $n = k$ $$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$

Show it is true for $n= k + 1$ $$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$ is a multiple of 24 $$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$ $$2\cdot7(24g + 5)3\cdot 5 - 5$$ I'm stuck and don't know how to proceed

You write $3\cdot 7^k$ in some places and $3\cdot 5^k$ in others. Which is it? — Wojowu, Dec 26 '19 at 21:30
Welcome to Mathematics Stack Exchange. Is it $3\cdot7^n$ or $3\cdot5^n$? — J. W. Tanner, Dec 26 '19 at 21:31
Your expression is simply $5(7^n-1)$. It is not divisble by $24$ when $n=3$. So there is a typo somewhere. — Anurag A, Dec 26 '19 at 21:33
Note that if $\gcd(a,24)=1$ then $a^2-1$ is a multiple of $24$ — J. W. Tanner, Dec 26 '19 at 21:33
It's suppose to be $3\cdot5^n$ I made the necessary corrections — Drex, Dec 26 '19 at 21:46
See https://math.stackexchange.com/questions/3486390/is-it-valid-to-subtract-in-the-middle-of-the-proof-here/3486393?r=SearchResults#3486393 — lab bhattacharjee, Dec 26 '19 at 22:27
Note: mod $8$ it's $2\cdot(-1)^n+3\cdot 5^n-5\equiv $ either $2+3\cdot1-5\equiv0 $ or $-2+3\cdot5-5\equiv 0$, and mod $3$ it's $2\cdot1^n-5\equiv0$ — J. W. Tanner, Dec 26 '19 at 23:18

score 2 · Answer 1 · answered Dec 26 '19 at 21:44

2

Hint:

$24 $ divides $2\cdot7^n+3\cdot5^n-5$ for $n=0$ and $1,$

and $2\cdot7^{n+2}+3\cdot5^{n+2}-5=49(2\cdot7^n+3\cdot5^n-5)-24(3\cdot5^n-10).$

answered Dec 26 '19 at 21:44

J. W. Tanner

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Could you further simplify this expression $49(2\cdot7^n+3\cdot5^n-5)-24(3\cdot5^n-10).$ – Drex Dec 26 '19 at 22:27
don't need to; it's a difference of two multiples of $24$ so it's a multiple of $24$, QED – J. W. Tanner Dec 26 '19 at 22:29
I mean how you expanded $2\cdot7^{n+2}+3\cdot5^{n+2}-5$ to $49(2\cdot7^n+3\cdot5^n-5)-24(3\cdot5^n-10).$ – Drex Dec 26 '19 at 22:33
oh, intermediate step is $49\cdot2\cdot7^n+(49-24)3\cdot5^n-(49-2\cdot24)5$ – J. W. Tanner Dec 26 '19 at 22:41
Thanks a lot, I now get it. Could you share any learning material in any form, whether pdf or online, where I could learn more. – Drex Dec 26 '19 at 22:53
to start with, look at answers here and here – J. W. Tanner Dec 26 '19 at 23:13

score 1 · Answer 2 · answered Dec 26 '19 at 21:46

Notice that $5^2$ and $7^2$ are both one more than a multiple of $24.$ Prove the cases $k=1$ and $k=2$ for your basis step.

Then in the induction you have, for one term,

$$2(7^{k+2}) = 2(7^k)(48+1).$$

Do the same for the other term. Induct by twos instead of by ones.

score 0 · Answer 3 · answered Dec 26 '19 at 21:56

In essence, the induction step relies on the identity $$(ax^n + by^n)(x+y) = (ax^{n+1} + by^{n+1}) + xy(ax^{n-1} + by^{n-1}).$$ You can verify this by simple multiplication. So if we define $$f_n(a,b,x,y) = ax^n + by^n,$$ then we have the recursion relation $$f_{n+1}(a,b,x,y) = (x+y)f_n(a,b,x,y) - xy f_{n-1}(a,b,x,y).$$ In your case, it is natural to choose $a = 2$, $b = 3$, $x = 7$, $y = 5$ to obtain $$f_{n+1} = 12 f_n - 35 f_{n-1}$$ (where I have omitted the arguments of the function). So if we let $g_n = f_n - 5$, then $$g_{n+1} + 5 = 12(g_n + 5) - 35(g_{n-1} + 5)$$ or $$g_{n+1} = 12g_n - 35 g_{n-1} - 120.$$ Since $120 = 5(24)$, it follows that $24 | g_{n+1}$ whenever $24 | g_n$ and $24 | g_{n-1}$; since $g_0 = 0$ and $g_1 = 24$, the proof is complete.

score 0 · Answer 4 · answered Dec 26 '19 at 21:56

First note that $5^k-5$ is divisible by $4$ for all positive integers $k$.

Let $P(n)$ be the proposition that $2\times7^n+3\times5^n-5$ is a multiple of $24$.

$P(1)$ is the proposition that $2\times7+3\times5-5$ is a multiple of $24$ which is true

Suppose $P(k)$ is true and so $2\times7^k+3\times5^k-5$ is a multiple of $24$.

Then $2\times7^{k+1}+3\times5^{k+1}-5=14\times 7^k+15\times5^k-5=7\times (2\times7^k+3\times5^k-5)-6\times(5^k-5)$ is divisible by $24$. Therefore $P(k+1)$ is true and we are finished.

score 0 · Answer 5 · answered Dec 26 '19 at 22:31

0

If induction is not mandatory,

I believe this is how the problem came into being

$$2(1+6)^n+3(1+4)^n\equiv2(1+6n)+3(1+4n)\pmod{(72,48)}$$

answered Dec 26 '19 at 22:31

lab bhattacharjee

279,016

Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$

5 Answers5