- #1
Math100
- 768
- 219
- Homework Statement
- Prove the assertion below:
The only prime p for which 3p+1 is a perfect square is p=5.
- Relevant Equations
- None.
Proof: Suppose that p is a prime and 3p+1=n^2 for some n##\in\mathbb{Z}##.
Then we have 3p+1=n^2
3p=n^2-1
3p=(n+1)(n-1).
Since n+1>3 for ##\forall n\in\mathbb{N}##,
it follows that n-1=3, and so n=3+1=4.
Thus 3p+1=n^2
=4^2
=16,
which implies that 3p=16-1=15,
this means p=15/3=5.
Therefore, the only prime p for which 3p+1 is a perfect square is p=5.
Then we have 3p+1=n^2
3p=n^2-1
3p=(n+1)(n-1).
Since n+1>3 for ##\forall n\in\mathbb{N}##,
it follows that n-1=3, and so n=3+1=4.
Thus 3p+1=n^2
=4^2
=16,
which implies that 3p=16-1=15,
this means p=15/3=5.
Therefore, the only prime p for which 3p+1 is a perfect square is p=5.