# We can divide $7^{17} - 7^{15}$ by? [closed]

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

The answer is $6$, but how?

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## closed as off-topic by Zachary Selk, Martin Sleziak, SchrodingersCat, Claude Leibovici, G. SassatelliDec 23 '15 at 9:09

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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... – pjs36 Dec 23 '15 at 5:32
@pjs36 I get that the congruences tag was probably irrelevant. – SchrodingersCat Dec 23 '15 at 5:35

Note that

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

and 48 is divisible by 6.

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@lam3r4370: JM is just factoring. – Jason S Sep 8 '10 at 13:21
7^17 - 7^15 = (7^2 - 1) * 7^15 = (49 - 1) * 7^15 = 48 * 7^15. No division required. – stevenvh Sep 8 '10 at 13:35
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 :-) – stevenvh Sep 8 '10 at 14:11
@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. – J. M. Sep 9 '10 at 0:26
@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: meta.stackexchange.com/questions/31253/… – Aryabhata Sep 10 '10 at 2:10

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. :) – J. M. Sep 8 '10 at 20:27
Could you please explain the factorization part of the example ? I am not getting how are you doing them. – Quixotic Nov 3 '10 at 7:12
Differences of squares e.g. x^4 + 2^2 = (x^2 + 2)^2 - (2x)^2 – Bill Dubuque Nov 3 '10 at 12:54