6
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We can divide $7^{17} - 7^{15}$ by?

The answer is $6$, but how?

Thanks in advance.

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    @Schrodingerscat I get that we're bumping old posts, but please try to choose tags more carefully. But you've got the hat, so hopefully that's that...2015-12-23
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    @pjs36 I get that the congruences tag was probably irrelevant.2015-12-23

2 Answers 2

19

Note that

$7^{17}-7^{15}=(49-1)*7^{15}$

and 48 is divisible by 6.

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    Yes ,but how did You get 49-1?If we divide 7^17 - 7^15 by 7^15 - yes ,but there isn't any division?Or I made mistake?2010-09-08
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    @lam3r4370: JM is just factoring.2010-09-08
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    Remember that we can use the properties of integer exponents here: $7^{17}=7^{15}7^2$, after which we can factor out what can be factored.2010-09-08
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    7^17 - 7^15 = (7^2 - 1) * 7^15 = (49 - 1) * 7^15 = 48 * 7^15. No division required.2010-09-08
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    I.e. 7^15*7^2-7^15 from where we get 1 ? 7^15-7^15? Sorry for the idiot questions ,but I don't understand it well.2010-09-08
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    $7^{17}-7^{15}=7^{15}\cdot 7^2-7^{15}\cdot 1=7^{15}(7^2-1)$2010-09-08
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    7^17 - 7^15 = 7^2 * 7^15 - 1 * 7^15 = (7^2 - 1) * 7^152010-09-08
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    Yes , I get it. 7^2 is 49.7^15*7^2-7^15 we just can imagine -7^15 as -1 ,because then we multiply it with 7^15 and it's the same.Thanks!2010-09-08
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    I hope this was a multiple choice question, because 6 is not the only answer. There's 1, 2, 3, 4, 6, 7, 8, 12, etcetera. 160 divisors in all :-)2010-09-08
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    Yes,you are right!2010-09-08
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    @J.M. Please do not answer questions like this in such great detail! This could easily be a hw question.2010-09-08
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    @BBischof, I'm not even sure what else could I have said that wouldn't be so vague, like "hint: use the law of exponents", and I'm not even sure that it would be comprehensible. So I took the least evil (IMHO) approach.2010-09-09
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    @J.M. I am quite fond of Bill's answer below. In general, his answers to this sort of question are fantastic. Almost always are they relevant, and helpful without giving it away completely. Additionally, he has a great nack for pointing out a more interesting way to look at the problem than I had considered. Taking a look at his answers is always enjoyable. However, I will say that on this one, it was a little bit trickier to not just give the answer. :/ I am not angry with what you did, just reminding you what we(sic) hope users will do.2010-09-09
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    @BBischof: Well, for one thing, I was hampered by the fact that I never knew "Aurifeuillean" until Bill brought it up. ;) And no, I never got the impression you were angry or anything like that; I'm just explaining what seemed expedient to me at the time of writing.2010-09-09
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    ok fair enough :)2010-09-09
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    @BBischof: The point of stack exchange sites is to give _answers_, not hints. If you think it is homework, clarify it before answering. Given the fact that OP even has this question (thus indicating a mathematics newbie), talking about various weirdly named polynomial identities, is very likely irrelevant to the OP, even as a well phrased hint. Of course, Bill's answer would definitely help others who happen to come across this question. On the other hand, those who can understand Bill's answer probably don't need any help on this question.2010-09-09
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    I've given more elaborate answers in this site, and the solution to an arithmetic problem grants me a "Nice Answer" badge. There is no justice in this world. XD2010-09-09
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    @Moron I think you are making too much of my comments, this sort of thing has been on discussed on meta extensively, this conversation need not persist here, and probably not anywhere.2010-09-10
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    @J.M: It happens. The upvotes are an indication of how the community understands/likes your answer, nothing to do about the difficulty/brilliance/quality of the answer. Perhaps this might be of interest to you: http://meta.stackexchange.com/questions/31253/the-bike-shed-problem-and-so2010-09-10
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    @BBischof: If you are talking about homework, I actually agree with you (of not just handing the solution on a platter). If you are talking about answers way beyond the asker's capabilities (like Bill's) then I don't see why there should be no conversations. Anyway, I will stop now.2010-09-10
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    @Moron's response to J.M. Nice link.2010-09-10
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    @Moron's response to me, I agree on both points.2010-09-10
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    Indeed, the meta discussion Moron linked to puts a lot of my observations on this site in perspective.2010-09-10
9

HINT $\rm\quad X^{n+2} - X^n \;=\; (X^2 - 1)\: X^n \;=\; (X-1)\: (X+1)\: X^n$

Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:

$$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\ \frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\ \end{array}$$

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    "Aurifeuillean"... thanks for posting this, it's a new word for me. :)2010-09-08
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    Could you please explain the factorization part of the example ? I am not getting how are you doing them.2010-11-03
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    Differences of squares e.g. x^4 + 2^2 = (x^2 + 2)^2 - (2x)^22010-11-03