prove that a perfect square can never be of the form 3k+2 for any value of k.?

Question

Please show ur working out properly:)

Answer 1

Make a proof by contradiction: suppose that there is an n such that n^2=3k+2. Decompose n as n=3p+q, with q being 0, 1 or 2 (rest of the division by 3).
Then 3p^2+6pq+q^2 = 3k+2 that I can rewrite as: 3(p^2+2q-k) = 2-q^2.
This implies that 2-q^2 should be a multiple of 3. But that's not the case for q=0, 1 or 2.

Answer 2

Every number must be of the form 3a+0, 3a+1 or 3a+2 for some a. All we're saying here is that a number must have a remainder of 0, 1 or 2 when divided by 3.

If a number of the form 3a+0 is squared, it gives another number of the form 9a² which is of the form 3b for some b. This can not be expressed as 3k + 2.

If a number of the form 3a+1 is squared, work out what form it must be in.

Do the same for 3a+2. It's easy, then, to show that none of them will give a number of the form 3k+2.

Answer 3

I'm assuming you mean any integer k, because if k=2/3, then 3k+2 = 3(2/3)+2 = 2+2 = 4.

Answer 4

Squrroot( k ) = 3k + 2
k = 9ksqur + 12k + 4
0 = 9ksqur + 11k + 4

Since everything is positive and the vertex starts on the 4 mark on the Y axis. It never touches the X axis when equivilent to 0. Thus, there is no solution.

Answer 5

could be forty 9, (x + 7) situations (x + 7) divide your b term in 0.5 and sq. it to return up with the respond. watch your signs and indicators additionally, those are all effective so its advantageous. If it reported x2 + 14x - ok then there does not be a real answer. n

Answer 6

What with induction?? Ha, I could NEVER do that. Good luck though.