- #1

Math100

- 783

- 220

- 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.