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I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows

$$2×7^n-2+3×5^n-3\\ 2(7^n-1)+3(5^n-1)\\ 2×6a+3×4b\\ 12(a+b)$$

In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.

Bernard
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aman rana
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  • If you want to use modular arithmetic to do it, perhaps it's useful to note that $7^2 \equiv 5^2 \equiv 1 \ (\text{mod} \ 24)$. – AlephNull Mar 20 '19 at 11:27

6 Answers6

5

Yes, it can be done by another method. Note that $7^2=2\times24+1$ and that $5^2=24+1$ and that therefore$$7^n\equiv\begin{cases}7\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise}\end{cases}$$and$$5^n\equiv\begin{cases}5\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise.}\end{cases}$$So:

  • if $n$ is odd, then $2\times7^n+3\times5^n-5\equiv2\times7+3\times5-5=24\equiv0\pmod{24}$;
  • otherwise, $2\times7^n+3\times5^n-5\equiv2\times1+3\times1-5\equiv0\pmod{24}$.
José Carlos Santos
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    may I suggest that you use $\times$ instead of the dot? – Ertxiem - reinstate Monica Mar 20 '19 at 11:29
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    Yes, you are allowed to do that. :-) – José Carlos Santos Mar 20 '19 at 11:30
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    Thank you very much. :) – Ertxiem - reinstate Monica Mar 20 '19 at 11:35
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    This is a correct method, but to prove the results stated would still require induction, for example to conclude that $7^{2k} \equiv 1 \pmod {24}$ from $7^2 \equiv 1 \pmod{24}$. Even proving that $1^n = 1$ will require induction. – Carl Mummert Mar 20 '19 at 12:10
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    @Carl Do you know any way to formalize "proof without induction" at an elementary level? – Bill Dubuque Mar 21 '19 at 02:49
  • @BillDubuque: You can classify according to the quantifier complexity of the induction. Most high-school 'induction' exercises use only $Δ_0$-induction, which is why most high-school students don't understand induction but think they do, as they cannot correctly do induction on a property involving unbounded quantifiers. – user21820 Dec 08 '19 at 09:20
  • @user21820 Yes, having studied logic I know related results, but by "elementary level" I meant at a level accessible to someone who has never studied logic - analogous to "naive" presentations of set theory, etc. – Bill Dubuque Dec 08 '19 at 15:54
  • @BillDubuque: I'm not sure what's the best way, but one could say induction on properties that can be computed by standard for-loops? Haha.. Otherwise one cannot catch all the stuff that could be argued to feel like 'no induction'. – user21820 Dec 08 '19 at 15:58
4

$$2(7^n-1)+3(5^n-1)$$

$$=2((1+6)^n-1)+3((1+4)^n-1)$$

$$\equiv2(6n+\text{ terms containing }6^2)+3(4n+\text{ terms containing }4^2)$$

$$\equiv 24n\pmod{24}$$

lab bhattacharjee
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Note that you have $$ 7^n - 1 = 6a\\ 5^n - 1 = 4b $$ Now we're interested in whether $a$ and $b$ are even or odd. Which is to say we want to know when $7^n - 1$ is divisible by $4$ (so that when you divide it by $6$ you get an even number), and when $5^n-1$ is divisible by $8$ (so that when you divide it by $4$, you get an even number).

The binomial theorem gives $$ 7^n - 1 = (8-1)^n - 1\\ = 8^n - \binom n18^{n-1} + \cdots + (-1)^{n-1}\binom{n}{n-1}8 + (-1)^n - 1 $$ We see that this is divisible by $4$ exactly when $(-1)^n - 1$ is, which is to say when $n$ is even.

Then we have $$ 5^n - 1 = (4 + 1)^n - 1\\ = 4^n + \binom n14^{n-1} + \cdots + \binom{n}{n-1}4 + 1 - 1 $$ and we see that this is divisible by $8$ precisely when $\binom{n}{n-1} = n$ is even.

So $a$ and $b$ are both even for even $n$, and both odd for odd $n$, proving that $a+b$ is always even, meaning $12(a+b)$ is divisible by $24$.

Arthur
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Case 1 : $n$ is odd

In this case $$2×7^n+3×5^n-5{=2×7^n+2×5^n+5^n-5\\=2\times\underbrace{(7^n+5^n)}_{12k}+5(5^{n-1}-1)\\=24k+5(\underbrace{25^{n-1\over 2}-1}_{24k'})\\=24k''}$$

Case 2 : $n$ is even

In this case $$2×7^n+3×5^n-5{=14×7^{n-1}+15×5^{n-1}-5\\=14\times\underbrace{(7^{n-1}+5^{n-1})}_{12k}+5(5^{n-2}-1)\\=24k+5(\underbrace{25^{n-2\over 2}-1}_{24k'})\\=24k''}$$

Mostafa Ayaz
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You may split it up by calculating $\mod 8$ and $\mod 3$:

  • $\mod 8$: \begin{eqnarray*} 2×7^n+3×5^n-5 & \equiv_8 & 2\times (-1)^n + 3\times (-3)^n +3 \\ & \equiv_8 & 2\times (-1)^n + 3((-3)^n + 1)\\ & \stackrel{3^2 \equiv_8 1}{\equiv_8}& \begin{cases} 2+3\times (1+1) & n = 2k \\ -2 +3 (-3 + 1) & n= 2k+1\end{cases}\\ & \equiv_8 & 0 \end{eqnarray*}
  • $\mod 3$: \begin{eqnarray*} 2×7^n+3×5^n-5 & \equiv_3 & 2\times 1^n + 3\times (-1)^n +1 \\ & \equiv_3 & 3\times (1 + (-1)^n)\\ & \equiv_3 & 0 \end{eqnarray*}
trancelocation
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$\!\!\!\!\!\begin{align} 2(7^n-1)&+3(5^n-1)\\ =\ 2×6a&\ +\ 3×4b\\ =\ &\color{#90f}{12}(\color{#0a0}{a+b})\\ \small \text{ but it is not enough [to prove divisibility by}\ & \color{#90f}{24}]\end{align}$

Finish simply with $\ \color{#c00}2\mid \color{#0a0}{a\!+\!b}\, =\, \dfrac{7^{\large n}\!-1}{7-1} + \dfrac{5^{\large n}\!-1}{5-1}\, =\, \overbrace{7^{\large n-1}\!+5^{\large n-1}}^{\rm\color{#c00}{even}}\! +\cdots + \overbrace{7+5}^{\rm\color{#c00}{even}}\, +\, \overbrace{1+1}^{\rm\color{#c00}{even}}$

Bill Dubuque
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  • Note induction is in fact used implicitly (in the proof of the lemma for the sum of a geometric series). This is true for all the answers - induction is used in some result (implicitly) invoked. It is not easy to formalize "without induction" at an elementary level. – Bill Dubuque Dec 08 '19 at 16:09