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I am trying to find if $$4^{1536} - 9^{4824}$$

is divisible by 35. I tried to show that it is not by finding that neither power is divisible by 35 but that doesn't entirely help me. I just know that I can't use fermats little theorem to help solve it.

  • It doesn't help, because both are clearly not divisible by 35. Just find if the two powers have the same residue mod 35 – Old John Feb 13 '15 at 23:40
  • It's definitely divisible by $5$ as its last digit is $5$ (check last digits of $4^n$ and $9^n$. Perhaps this might be useful (so you need to check divisibility by $7$). – Andrei Rykhalski Feb 13 '15 at 23:41
  • @AndreiRykhalski: $4^4-9^1=247$ is not divisible by $5$. – barak manos Feb 13 '15 at 23:45
  • @barakmanos I didn't mean that $4^n$ - $9^m$ for all $n$ and $m$, but the fact that last digit of a power of a number is a periodic function. – Andrei Rykhalski Feb 13 '15 at 23:48

6 Answers6

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First, $$ 4^{1536}-9^{4824}\equiv(-1)^{1536}-(-1)^{4824}\equiv 1-1\equiv 0\pmod{5}. $$ Second, $$ 4^{1536}-9^{4824}=64^{512}-729^{1608}\equiv 1^{512}-1^{1608}\equiv 1-1\equiv 0\pmod{7}. $$

Edit: if you are unfamiliar with modular arithmetic, think in terms the Binomial Theorem. For example, (below, $A$ is some integer) $$ 4^{1536}=(5-1)^{1536}=5A+(-1)^{1536}=5A+1 $$ where the last equality follows because $1536$ is even.

yurnero
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  • How does 4 turn into -1? –  Feb 13 '15 at 23:57
  • $4\equiv-1\pmod{5}$ because $4-(-1)=5$ is divisible by $5$. – yurnero Feb 13 '15 at 23:58
  • I don't follow that logic, can you demonstrate with a more simple example? –  Feb 14 '15 at 00:16
  • I don't know the binomial theorem either but how is it that you are taking the powers of 4 and 9 and turning them both into 1? –  Feb 14 '15 at 00:28
  • @user138246 $a\equiv b\pmod{p}\implies a^n\equiv b^n\pmod{p}, \forall n\in\mathbb N$, because $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1})$, so $4\equiv -1\pmod{5}\implies 4^{1536}\equiv (-1)^{1536}\equiv 1\pmod{5}$, because $(-1)^{1536}=1$. Same logic for everything else. – user26486 Feb 14 '15 at 00:38
  • Where id 64 come from? –  Feb 14 '15 at 02:00
  • @user138246 $4^{1536}-9^{4824}\equiv (4^{3})^{512}-(9^{3})^{1608}\equiv 64^{512}-729^{1608}\pmod{7}$. – user26486 Feb 14 '15 at 02:05
  • Yes I see that but how did you figure out to extract the numbers in that manner? –  Feb 14 '15 at 02:09
  • @user138246 I played around with small powers of $4$ and $9$ so that the resulting numbers are $1$ or $-1$ away from multiples of $7$. After a few trials, I realized that $4^3$ and $9^3$ would work. Hope this helps. Good luck! – yurnero Feb 14 '15 at 02:13
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Note $4^6\equiv (-6)^2 \equiv 1$ (mod $35$) and $6 \mid 1536$, so $4^{1536}\equiv 1$ (mod $35$).

Similarly $9^6 \equiv 1$ (mod $35$) $\implies 9^{4824}\equiv 1$ (mod $35$).

So we can conclude that $4^{1536}-9^{4824} \equiv 0$ (mod $35$), i.e. the number is divisible by $35$.

Alex
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  • I don't understand the $6 | 1536$ part, what does that mean? –  Feb 14 '15 at 00:19
  • @user138246 $6\mid 1536\implies 1536=6c$ for some $c\in\mathbb N$, so $4^{1536}\equiv 4^{6\cdot c}\equiv (4^6)^c\equiv 1^c\equiv 1\pmod{35}$. We had $(4^6)^c\equiv 1^c\pmod{p}$ because $4^6\equiv 1\pmod{p}$ and $a\equiv b\pmod{p}\implies a^n\equiv b^n\pmod{p},\forall n\in\mathbb N$, because $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$. – user26486 Feb 14 '15 at 00:43
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Hint mod $\,5\!:\ 4\equiv -1\equiv 9\,\Rightarrow\, \color{#c00}{4^6\equiv 1\equiv 9^6}$

and, $ $ mod $\,7\!:\ 9^3\equiv 2^3\equiv 1\,\Rightarrow\ \color{#c00}{4^6\equiv 1\equiv 9^6}$

So mod $\,35\!:\ \color{#c00}{4^3\equiv 1\equiv 9^3}\ $ by CRT $ $ (or by $\,5,7\mid a^6\!-\!1\,\Rightarrow\,{\rm lcm}(5,7)=35\mid a^6\!-\!1)$

So mod $\,35\!:\ \color{#c00}4^{\color{#c00}3j} - \color{#c00}9^{\color{#c00}9^k}\equiv \color{#c00}1^{j}-\color{#c00}1^{k} \equiv 1$

In particular it's true if the exponents are even with digit sum divisible by $\,3,\,$ as in your case.

Remark $\ $ Since you say congruence arithmetic is unfamiliar, below is a more elementary proof using only that $\,a-b\mid a^n-b^n.\ $ Suppose $\ 2\mid j\!-\!i \ge 0.\ $ Then

$\qquad\quad \begin{align} 9^{3j}-4^{3i} = &\ 9^{3j}-4^{3j}\ +\ 4^{3j}-4^{3i}\\ = &\ \color{#0a0}{9^{3j}-4^{3j}}\ +\ 4^{3i}\,(\color{#c00}{4^{3(j-i)}\!-1})\end{align}$

But $\ 5,7\mid 9^3-4^3\mid \color{#0a0}{9^{3j}-4^{3j}}$ and $\,5,7\mid 4^6-1\mid \color{#c00}{4^{3(j-i)}\!-1}\,$ by $\,2\mid j\!-\!i\,\Rightarrow\, 6\mid 3(j\!-\!i) $

Thus $\,5,7\,$ also divide their sum, therefore their lcm = $\,5\cdot 7 = 35\,$ divides their sum.

Bill Dubuque
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  • Can you explain your notation, I am not familiar with it enough to read this commit. I don't know what the : implies, I don't know what CRT is, I don't know what lcm is either. Thanks. –  Feb 14 '15 at 00:15
  • @user138246 Do you know congruence arithmetic? i.e. $,a\equiv b\pmod m\ $ iff $\ m\mid a-b\ \ $ – Bill Dubuque Feb 14 '15 at 00:21
  • I do not and I do not know what | means –  Feb 14 '15 at 00:22
  • @user138246 Usually these problems are posed after one learns a bit of elementary number theory. You should state in your question that you are not familiar with congruences or modular arithmetic. What textbook are you using? – Bill Dubuque Feb 14 '15 at 00:29
  • I know a little bit of modular arithmetic and the idea of congruence, I am just not familiar with the math heavy notation and acronyms. –  Feb 14 '15 at 00:30
  • See the Congruence Arithmetic Rules, e.g. above I used that $,{\rm mod}\ 5!:\ \color{#0a0}{4\equiv -1},\Rightarrow, \color{#0a0}4^6\equiv (\color{#0a0}{-1})^6\equiv 1,$ by the Congruence Power Rule. – Bill Dubuque Feb 14 '15 at 00:33
  • Why are you using negative numbers in modulo arithmetic? Why does it matter if a number if negative? –  Feb 14 '15 at 00:35
  • @user138246 Because it is a lot easier to compute $,(-1)^n$ than $,4^n,,$ esp. for very large $,n,,$ as in your problem. By the congruence arithmetic rules we are allowed to replace arguments of sums and products (but not exponents!) by any congruent integer. Usually we choose one of smallest magnitude so to simplify the arithmetic. – Bill Dubuque Feb 14 '15 at 00:39
  • so if we are working with mod 3 then 1, 4, 7, 11 and so on are all replaceable regardless of their power? –  Feb 14 '15 at 00:40
  • @user138246 $ $ Yes, $,7\equiv 4\equiv 1,\Rightarrow, 7^n\equiv 4^n\equiv 1^n\equiv 1\pmod{3},$ by the $ $ Congruence Power Rule, $ $ (a power is an iterated product, and it is legal to replace the arguments of products). – Bill Dubuque Feb 14 '15 at 00:43
  • @user138246 I added a simpler proof in a remark. Here $, j\mid k,$ means $,j,$ divides $,k,,$ i.e. $,k/j,$ is an integer. – Bill Dubuque Feb 14 '15 at 01:03
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Fermat's little theorem:

"If $p$ is a prime number, then: $\forall$ $x \in \Bbb N / gcd(x,p) = 1$, $x^{p-1} \equiv 1 \pmod p$".

Let $N = 4^{1536} - 9^{4824}$.

$4^{1536} = (4^6)^{256} \equiv 1 \pmod 7$ (Fermat's little)

$9^{4824} = (9^6)^{804} \equiv 1 \pmod 7$ (Again, Fermat)

Thus: $N \equiv 1 - 1 \pmod 7 \equiv 0 \pmod 7$. This shows that $7 | N$.

Now:

$4^{1536} = (4^4)^{384} \equiv 1 \pmod 5$ (Fermat..)

$9^{4824} = (9^4)^{1206} \equiv 1 \pmod 5$ (Again)

Thus: $N \equiv 1 - 1 \pmod 5 \equiv 0 \pmod 5$.

This shows $5|N$.

Therefore $5|N$ and $7|N$. This gives $35|N$.

  • Why is $4^{6*256} = 1 (mod 7)$ I don't follow. 1 mod 7 is just 1 but how did you conclude that 4 raised to some massive power is equal to 1? It makes no sense to me. –  Feb 13 '15 at 23:59
  • $4^6 \equiv 1 \pmod 7$ using the aforementioned theorem. Now raise both sides to that "massive power". $1$ will still be $1$. –  Feb 14 '15 at 00:01
  • Yes I see that but you are making the claim that some number (that is not 1) is equal to 1. I do not see how that stands. –  Feb 14 '15 at 00:04
  • "$a \equiv b \pmod n$" means that the remainder of division of $a$ by $n$ is $b$. It doesn't mean that $a = b$. –  Feb 14 '15 at 00:08
  • Was the "I can't use Fermat" bit present from the beginning? I completely missed that. –  Feb 14 '15 at 00:11
  • It was there but I meant that I tried to use it and was unable. –  Feb 14 '15 at 00:14
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I have little background in this area, but I can determine the answer, ergo anyone else with little background should have no trouble following along.

I know that $x^n$ mod $p$ will follow a cyclic pattern as n is iterated, so I'll first find what those cycles are for $4^n$ mod $35$ and $9^n$ mod $35$, for $n = 0,1,2,3...$

$4^n$ mod $35 = 1,4,16,29,11,9,1,4,16,29...$

$9^n$ mod $35 = 1,9,11,29,16,4,1,9,11,29...$

So we see that both cycles have a length of 6, so you can know what each number's $nth$ power mod $35$ is by what the exponent $n$ mod $6$ is. $4^a - 9^b$ will be divisible by $35$ if $4^a$ and $9^b$ have the same value mod $35$. Importantly, the two cycles above are in fact the reverse of each other. So $4^a$ will have the same value mod $35$ as $9^b$ if $a$ mod $6 = (6 - (b$ mod $6))$ mod $6$. As it happens, both $1536$ mod $6$ and $4824$ mod $6$ equal $0$, so they do satisfy the aforementioned equation, and therefore $4^{1536} - 9^{4824}$ IS divisible by $35$.

Alan Rat
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I will suggest to use Euler's Theorem which states $\forall a \in \mathbb{Z}$ s.t. gcd$(a, n)=1$, then $a^{\phi(n)} \equiv 1$ mod $n$

In your case we have $n=35$, $\quad a=4,9\quad$ and $\quad \phi(35)=24$

So $\quad \quad 4^{24} \equiv 1$ mod $35\quad$ and $\quad 9^{24} \equiv 1$ mod $35\quad$

Since $\quad4$x$64 = 1536 \quad$ and $\quad 9$x$201 = 4824$

So $\quad \quad 4^{1536} \equiv 1$ mod $35\quad$ and $\quad 9^{4824} \equiv 1$ mod $35\quad$

Hence $\quad 4^{1536} - 9^{4824} \equiv 1 - 1 \equiv 0$ mod $35\quad \implies 35 | 4^{1536} - 9^{4824}$

user110219
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