- #1
Math100
- 768
- 219
- 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?