Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
estudier
Group Title
Show that any number that is both square and cube (eg 64) is of form 7k or 7k +1
 one year ago
 one year ago
estudier Group Title
Show that any number that is both square and cube (eg 64) is of form 7k or 7k +1
 one year ago
 one year ago

This Question is Closed

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i assume we cannot consider 0
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Note that any number must be 7k + r with r = 0,1,...6
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
@estudier i think all numbers can be expressed that way
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
(7k + r)^3 = (7k + r)^2 would that help  i'm clutching at straws to be honest!!
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
So I think it is 7k+1
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
"(7k + r)^3 = (7k + r)^2" Maybe expand these instead of setting them equal...
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
1^6=7*0+1 2^6=7*9+1 3^6=7*104+1 4^6=7*585+1 I mean every number can be expressed as 7k + r with r = 0,1,...6
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
So, I think u mean to say 7k+1
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
I mean every number can be expressed as 7k + r with r = 0,1,...6 Yes, I said that above....
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
oh.....THAT WAS A HINT
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
The question does say 7k or 7k +1
 one year ago

NewbieCarrot Group TitleBest ResponseYou've already chosen the best response.0
For 7k, (7k)^2 (mod 7)= 0^2 (mod 7)=0, (7k)^3 (mod 7)=0^3 (mod 7)=0 For 7k+1, (7k+1)^2 (mod 7)= 1^2 (mod 7)=1, (7k+1)^3 (mod 7)= 1^3 (mod 7)=1 For 7k+2, (7k+2)^2 (mod 7)= 2^2 (mod 7)=4, (7k+2)^3 (mod 7)= 2^3 (mod 7)=1 ......... For 7k+6, (7k+6)^2 (mod 7)= 6^2 (mod 7)=1, (7k+6)^3 (mod 7)= 6^3 (mod 7)=6 for any 7k+r, if the result is r, then it has the same form.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
@NewbieCarrot Considering certain remainders is a good way to go....
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Well x=n^6...... Now, If n=0 MOD 7 then , n^6=0 MOD 7 Thus, x can be expressed as 7k.... Now if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7 > BUT I HAVEN'T PROVED THIS PART.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
(7k+r)^2 = 49k^2 + 14kr + r^2 = 7(7k^2+2kr) + r^2
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
(7k+r)^3 = 343k^3 +147k^2r +21kr^2 +r^3 = 7(49k^3 +21k^2r +3kr^2) +r^3
 one year ago

NewbieCarrot Group TitleBest ResponseYou've already chosen the best response.0
@estudier You got it.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Not done yet....
 one year ago

NewbieCarrot Group TitleBest ResponseYou've already chosen the best response.0
7(7k^2+2kr) + r^2 7(49k^3 +21k^2r +3kr^2) +r^3 have the same form with 7k+r while r=r^2=r^3 (mod 7)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
OK, the reaminders are the same, and.....?
 one year ago

NewbieCarrot Group TitleBest ResponseYou've already chosen the best response.0
The only possible answer is r=0 or 1. And we are done
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.1
n = 7k+r n^2 = 7x + r^2 n^3 = 7y + r^3
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.1
r = [1,6]
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
"The only possible answer is r=0 or 1. And we are done" True, but a little more explanation would be nice...:)
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.1
let,n = 7k+r (r = 1,2,3,4,5,6) n^2 = 7x + r^2 n^3 = 7y + r^3 plugin r = 1,2,3,4,5,6 and the intersection of remainders gives the form of a number thats both square and cube
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Yes, that's right, well done.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Just include r= 0 as well....
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.1
oh yes... .
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Is there this rule: If n=x MOD y then, n^a = x^a MOD y
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Is there this rule: If n=x MOD y then, n^a = x^a MOD y Yes (r>=1)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
GREAT....... Then I think I proved if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
a^(p1) = 1 mod p (gcd a,p = 1)
 one year ago

NewbieCarrot Group TitleBest ResponseYou've already chosen the best response.0
Here is my experience, when you see this sentence "is of form 7k or 7k +1" you always should try all 7k+r to get the results. Same as 3k.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
a^(p1) = 1 mod p (p prime, gcd a,p = 1) (OR a^p = a mod p) Fermat's Little Theorem
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
@sauravshakya
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Indirectly related to the "necklace2 problem....
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
WHAT!
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
But how???? That is related to permutation and combination.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
n^pn strings involving 2 or more colours are partitioned into disjoint sets of p strings each set being strings obtained from each other by by a sequence of cycles ie p divides n^pn which is Fermat's Little Theorem
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Special case of your necklace problem.... Loads of beads of n colours, how many different necklaces of p beads can be made, p is prime?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
(n^pn)/2p + [n(p+1/2) n]/2 + n
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Did u mean that every colour has only one bead.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
You make a string of p beads and join the ends, so that's n^p strings. Of those, n strings are beads of 1 colour alone (one for each colour) So n^pn strings having at least 2 colours etc etc....
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.