algebra question (squares?)

• Jan 20th 2009, 01:26 AM
requal
algebra question (squares?)
1 Write down the first ten square numbers (perfect squares) [can do this]
2 Show that when each of these squares are divided by 3, the remainder is either 0 or 1
3 Write down three consecutive numbers the first of which is divisible by three[can do up to this part]
4 Prove that the square of any number is of the form of 3n or 3n+1
5 Use modular arithmetic(what's that) to prove the same result
• Jan 20th 2009, 03:43 AM
Mush
Quote:

Originally Posted by requal
1 Write down the first ten square numbers (perfect squares) [can do this]
2 Show that when each of these squares are divided by 3, the remainder is either 0 or 1
3 Write down three consecutive numbers the first of which is divisible by three[can do up to this part]
4 Prove that the square of any number is of the form of 3n or 3n+1
5 Use modular arithmetic(what's that) to prove the same result

In part 2 you showed that when a square is divided by 3, the remainder is either 0 or 1.

This means that the number can be written in the form:

k = 3n+1
or
k = 3n.
• Jan 20th 2009, 11:16 PM
requal
Okay i got the worked solutions of the question, but I don't understand it, someone please clarify it for me.

iii, 3n, 3n+1, 3n+2
iv Any number must be of the form of 3k, 3k+1, 3k+2

Hence, \$\displaystyle (3k)^2=9k^2=3(3k^2)=3n\$
And, \$\displaystyle (3k+1)^2=9k^2+6k+1=3(3k+2)+1=3n+1\$
And, \$\displaystyle (3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1=3n+1\$

Can someone explain to me how they got the solution??(Angry) I mean how can n is equal to three randomly different terms
• Jan 21st 2009, 12:05 AM
Isomorphism
Quote:

Originally Posted by requal
Can someone explain to me how they got the solution??(Angry) I mean how can n is equal to three randomly different terms

Well Its another way of saying every number when divided by 3 leaves a remainder of either 0,1 or 2.