Division Algorithm help.?

Question

Ok, so i have a few problems, i need as much detail as possible. I'm having a hard time with my Division Algorithms.

1) Prove that if a and b are integers, with b>0, then there exist unique integers q and r satisfying a=qb+r, where 2b<(or equal to) r<3b

2) Prove that 3a^2-1 is never a perfect square. [hint: THe quare of any interger is either in the form of 3k or 3k+1 (Expain if you can)]

3) If n is an odd interger, show that n^4+4n^2+11 is in the form of 16k.

You can do any or all of these, i just ask that you explain it the best you can, as clear as you can. And maybe if any of y'all have the time would you mind attepting.... Prove that no interger in the following sequence is a perfect square: 11,111,1111,11111,....... [Hint: A typical term 111...111 can be written as 111...111=111...108+3=4k=3] <---I have no idea on this one, but if you wouldn't mind....

Thank you in advanced.

Answer 1

1) By division algorithm there are unique integers x,y satisfying a=xb + y, where 0&lt;= y &lt; b. So a=(x-2)b + 2b +y and 2b &lt;= 2b+y &lt;3b. So q=x-2 and r=2b +y are the needed integers. They are unique because if

a=ub + w where 2b&lt;= w &lt; 3b then a=(u+2)b + (w-2b) and

0&lt;= (w-2b) &lt;3b, which means x=u+2 and y=w-2b because x and y are unique, which means u=x-2=q and w=2b+y=r.

2)By division algorithm any integer is of the form 3k, 3k+1, or 3k +2.

Now (3k)^2 = 9k^2 =3(3k^2);  (3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1;

(3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1. So a perfect square is always of the form  3k or 3k+1, never 3k+2. But 3a^2-1=3(a^2-1)+2 is of the form 3k+2, so it is not a perfect square.

3)If n is odd then n=2k+1 for some integer k. So

n^4+4n^2+11=(2k+1)^4+4(2k+1)^2+11

=(16k^4+32k^3+24k^2+8k+1)+4(4k^2+4k+1) +11

=16k^4+32k^3+40k^2+24k+16

=16(k^4+2k^3+2k^2+k+1)+8k(k+1)

=16(k^4+2k^3+2k^2+k+1) + 16k(k+1)/2

k(k+1) is always even so n^4+4n^2+11 is of the form 16a for some integer a

4)111...1100 is divisible by 4 (because it is divisible by 100). So 111...1108 is of the form 4k. So 111...111 is of the form 4k+3. Use the technique in 2) to show that no perfect square is of the form 4k+3. So no 111...111 is a perfect square.

Hope this helps. Good luck!