How do I prove $19\cdot8^n+17$ is a composite number?
Or is that number just a prime?
So I tried to find a divisor in the cases $ n = 2k $ and $ n = 2k + 1 $. But I had no success.
Do you have any ideas?
Asked
Active
Viewed 590 times
3
-
Hint $\ $ Do a covering congruence argument as in this answer. – Bill Dubuque Jun 17 '15 at 15:03
-
@BillDubuque Hagen's answer is essentially a duplicate of your answer there. – user26486 Jun 17 '15 at 15:39
-
@user26486 Indeed. I didn't notice before that they are exactly the same numbers. Readers may find of interest the paper of Schinzel linked there. – Bill Dubuque Jun 17 '15 at 15:42
2 Answers
12
Three cases:
$n$ is even. Then $$19\cdot 8^n+17\equiv 1\cdot (-1)^n+17\equiv 1+17\equiv 0\pmod{\! 3}$$
$n=4k+1$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+1}+17\equiv 6\cdot \left(8^2\right)^{2k}\cdot 8+4\equiv 6\cdot (-1)^{2k}\cdot 8+4$$
$$\equiv 48+4\equiv 52\equiv 0\pmod{\! 13}$$
- $n=4k+3$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+3}+17\equiv (-1)\cdot \left(8^2\right)^{2k}\cdot 8^3+17\equiv (-1)\cdot (-1)^{2k}\cdot 8^3+17$$
$$\equiv -(-2)^3+17\equiv 8+17\equiv 25\equiv 0\pmod{\! 5}$$
user26486
- 11,331
-
1Impressive, but can you indicate how you came up with these divisors, and with (mod 4) for the cases? – alexis Jun 17 '15 at 13:16
-
1@alexis Hagen clearly shows the reasoning you can use to come up with this solution. – user26486 Jun 17 '15 at 13:29
-
9
Note that $8^4\equiv 1\pmod{3^2\cdot 5\cdot 7\cdot 13}$. Thus if you find a divisor $\in\{3,5,7,13\}$ for $n=r$, then the same divisor works for $n=4k+r$. So check $r=0,1,2,3$ for these divisors. (Clearly, $19\cdot 8^n+17>13$ so these divisors are proper).
Hagen von Eitzen
- 374,180