Remainder when 29 divides $3^{2002} + 7^{2002} + 2002$

Question

My attempt:

We note that-

$2002 \equiv 1$ (mod $29$)

$3^{2002}\equiv 3^{14}$ (mod $29$)

$7^{2002}\equiv 7^{14}$ (mod $29$) [From Fermat]

Now, how do I reduce $3^{14}$ and $7^{14}$?

Maybe you can use $7^{14} = 49^{7}$ and a similar identity for $3^{14}$ — Ruben, Jul 04 '19 at 16:24
I mean, what did you try? Cleary $7^{14} = 49^7 \equiv 20^7 \mod 29$. Now just continue reducing. — Ruben, Jul 04 '19 at 16:26
Iterated squaring isn't too bad. For $3^{14}$ you can remark that $3^3\equiv -2$ so $3^{12}\equiv 16$ and so on. If you just play with the exponents you can compute these things very rapidly. — lulu, Jul 04 '19 at 16:27
I believe that this method will involve a fair bit of tedious calculations? — Anurag Saha, Jul 04 '19 at 16:34

lab bhattacharjee · Accepted Answer · 2019-07-04T16:56:43.397

7

$$3^{2002}+7^{2002}=(3^2)^{1001}+(7^2)^{1001}$$ is divisible by $3^2+7^2$

edited Jul 04 '19 at 16:56

answered Jul 04 '19 at 16:32

lab bhattacharjee

See also : https://math.stackexchange.com/questions/909973/how-to-prove-that-53103-10353-is-divisible-by-39 https://math.stackexchange.com/questions/1819423/find-3333-7777-pmod-50/1819426#1819426 https://math.stackexchange.com/questions/253005/520121-is-divisible-with-313 – lab bhattacharjee Jul 04 '19 at 16:36
you have written $5^{2002}$ by mistake – Anurag Saha Jul 04 '19 at 16:42
@Anurag, Thanks – lab bhattacharjee Jul 04 '19 at 16:57

score 1 · Answer 2 · answered Jul 05 '19 at 01:54

1

To answer your original question, "Now, how do I reduce $3^{14}$ and $7^{14}$''

One approach would be (should you forget the theorem of a previous answer):

$$ 3^{14} \equiv (3^{3})^{4}(3^{2}) \equiv (-2)^{4}(9) \equiv -1 \pmod{29} $$

$$ 7^{14} \equiv (-9)^{7} \equiv (81)^{3}(-9) \equiv (-6)^{3}(-9) \equiv 1 \pmod{29}. $$

And so, your original number is divisible by $29.$

answered Jul 05 '19 at 01:54

DDS

1

Or $\ \large 7\equiv 6^{\Large 2},$ so $\large ,7^{\Large 14}\equiv 6^{\Large 28}\equiv 1\ $ by Fermat. Or we can use quadratic reciprocity. – Bill Dubuque Jul 05 '19 at 03:26
@BillDubuque . Or $7^2\equiv -9 \implies$ $ 7^4\equiv -6\implies $ $7^8\equiv 7\implies$ $ 7^7\equiv 1. $ Hence $7^{2002}=(7^7)^{286}\equiv 1 . $ – DanielWainfleet Jul 05 '19 at 03:37
1

@Daniel But already $\ \large 7^{\Large 2}\equiv -3^{\Large 2}\ $ to power $\ \large ,7,\Rightarrow, 7^{\Large 14}\equiv -3^{\Large 14}\ \ $ – Bill Dubuque Jul 05 '19 at 03:50

2 Answers2