- #1

Math100

- 756

- 205

- Homework Statement
- If p##\geq##5 is a prime number, show that p^{2}+2 is composite.

[Hint: p takes one of the forms 6k+1 or 6k+5.]

- Relevant Equations
- None.

Proof: Suppose p##\geq##5 is a prime number.

Applying the Division Algorithm produces:

p=6k, p=6k+1, p=6k+2, p=6k+3, p=6k+4 or p=6k+5 for some k##\in\mathbb{Z}##.

Since p##\geq##5 is a prime number,

it follows that p cannot be divisible by 2 or 3 and p must be odd.

Thus, p takes one of the forms 6k+1 or 6k+5.

Now we consider two cases.

Case #1: Suppose p=6k+1.

Then we have p^{2}+2=(6k+1)^{2}+2

=36k^{2}+12k+3

=3(12k^{2}+4k+1)

=3m,

where m=12k^{2}+4k+1 is an integer.

Thus, p^{2}+2 is composite.

Case #2: Suppose p=6k+5.

Then we have p^{2}+2=(6k+5)^{2}+2

=36k^{2}+60k+27

=3(12k^{2}+20k+9)

=3m,

where m=12k^{2}+20k+9 is an integer.

Thus, p^{2}+2 is composite.

Therefore, if p##\geq##5 is a prime number, then p^{2}+2 is composite.

Above is my proof for this problem. Can anyone please verify/review to see if this is correct?

Applying the Division Algorithm produces:

p=6k, p=6k+1, p=6k+2, p=6k+3, p=6k+4 or p=6k+5 for some k##\in\mathbb{Z}##.

Since p##\geq##5 is a prime number,

it follows that p cannot be divisible by 2 or 3 and p must be odd.

Thus, p takes one of the forms 6k+1 or 6k+5.

Now we consider two cases.

Case #1: Suppose p=6k+1.

Then we have p^{2}+2=(6k+1)^{2}+2

=36k^{2}+12k+3

=3(12k^{2}+4k+1)

=3m,

where m=12k^{2}+4k+1 is an integer.

Thus, p^{2}+2 is composite.

Case #2: Suppose p=6k+5.

Then we have p^{2}+2=(6k+5)^{2}+2

=36k^{2}+60k+27

=3(12k^{2}+20k+9)

=3m,

where m=12k^{2}+20k+9 is an integer.

Thus, p^{2}+2 is composite.

Therefore, if p##\geq##5 is a prime number, then p^{2}+2 is composite.

Above is my proof for this problem. Can anyone please verify/review to see if this is correct?