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anonymous
 4 years ago
Show that any number that is both square and cube (eg 64) is of form 7k or 7k +1
anonymous
 4 years ago
Show that any number that is both square and cube (eg 64) is of form 7k or 7k +1

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i assume we cannot consider 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Note that any number must be 7k + r with r = 0,1,...6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@estudier i think all numbers can be expressed that way

cwrw238
 4 years ago
Best 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!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I think it is 7k+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"(7k + r)^3 = (7k + r)^2" Maybe expand these instead of setting them equal...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01^6=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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, I think u mean to say 7k+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean every number can be expressed as 7k + r with r = 0,1,...6 Yes, I said that above....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh.....THAT WAS A HINT

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The question does say 7k or 7k +1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@NewbieCarrot Considering certain remainders is a good way to go....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(7k+r)^2 = 49k^2 + 14kr + r^2 = 7(7k^2+2kr) + r^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(7k+r)^3 = 343k^3 +147k^2r +21kr^2 +r^3 = 7(49k^3 +21k^2r +3kr^2) +r^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@estudier You got it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.07(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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK, the reaminders are the same, and.....?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The only possible answer is r=0 or 1. And we are done

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n = 7k+r n^2 = 7x + r^2 n^3 = 7y + r^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"The only possible answer is r=0 or 1. And we are done" True, but a little more explanation would be nice...:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let,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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, that's right, well done.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just include r= 0 as well....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is there this rule: If n=x MOD y then, n^a = x^a MOD y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is there this rule: If n=x MOD y then, n^a = x^a MOD y Yes (r>=1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0GREAT....... Then I think I proved if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a^(p1) = 1 mod p (gcd a,p = 1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a^(p1) = 1 mod p (p prime, gcd a,p = 1) (OR a^p = a mod p) Fermat's Little Theorem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Indirectly related to the "necklace2 problem....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But how???? That is related to permutation and combination.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n^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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Special case of your necklace problem.... Loads of beads of n colours, how many different necklaces of p beads can be made, p is prime?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(n^pn)/2p + [n(p+1/2) n]/2 + n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did u mean that every colour has only one bead.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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....
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