Question

which of the following statements is (are) true for all positive integers 1) n^3+2n is divisible by 3
2) 5^2^n-1 is divisible by 24

Answer 1

the first one might help to factor

Answer 2

if a|b , then b = ak for some integer k

b - ak = 0

Answer 3

i just need to know if they both are true or just one or if there false

Answer 4

you find that out by actually working thru the process; we are not running a free answering service here

Answer 5

if you can find a counterexample, that tends to be easiest to prove something false

Answer 6

its hard to parse the second one without some brackets or some such

Answer 7

the 2 and n are both squared

Answer 8

\[\Large 5^{2^{n-1}}\]

Answer 9

the 1 isnt

Answer 10

\[\Large 5^{(2^n-1)}\]

Answer 11

\[\Large 5^{2^{n}}-1\]

Answer 12

yea

Answer 13

\[n^3+2n=0~(mod~3)\]
\[n(n^2+2)=0~(mod~3)\]
well, if n is a multiple of 3 that works out fine, so id say we would need to determine if n^2+2 is always a multiple of 3

Answer 14

\[n^2+2=0~(mod~3)\]
\[n^2=1~(mod~3)\]
hmmm
this would indicate to me that for any given n, it would be its own inverse

Answer 15

* 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1

according to a Z3 multipication table of equivalence classes, that seems to fit the bill

Answer 16

so they both would be true

Answer 17

ive only tested the first one, and i believe it is true

Answer 18

the question remains:\[5^{2^n}=1~(mod~24)\]

Answer 19

this is really asking if \[5^k=1~(mod~24)\]

Answer 20

nope its asking

Answer 21

thats what i wrote .....

Answer 22

you said equals

Answer 23

\[\frac{5^{2^n}-1}{24}~\to~ 5^{2^n}-1={24k}\]
\[ 5^{2^n}={24k}+1~~\to~~5^{2^n}=1~(mod~24)\]

Answer 24

so true ?

Answer 25

so there both true

Answer 26

thats just moving things around, im not sure if its true yet or not; what do you have to bring to the table?

Answer 27

you dont have to be rude i didnt ask you to look at it i came on here for help bc i didnt know it ..thanks for the help .. i mean for what you did any ways

Answer 28

when n=1; 2^1 = 2, 5^2 = 25 = 24(1) + 1

when n=2; 2^2 = 4, 5^4 = 625 = 24(6) + 1

5^8 = 24(16276) + 1


so at a cursory glance id say its true as well .... but i cant be certain of that

Answer 29

im not being rude, i simply told you that i hadnt figured it out yet, and asked what ideas you had to offer ....

Answer 30

hmm, since 2^n is always an even number 0,2,4,8,16,32, ....

if 5^2 is congruent to 1 mod 24, then any multiple of it would be as well

Answer 31

\[5^2=1~(24)\]
\[5\cdot 5=25=1~(24)\]
\[(5^2)^n=(5\cdot 5)\cdot(5\cdot 5)\cdot(5\cdot 5)\cdot\cdot\cdot(5\cdot 5)\]
\[(5^2)^n=1\cdot1\cdot1\cdot\cdot\cdot1=1~(24)\]
so id say yes, its true as well