# The only prime p for which 3p+1 is a perfect square is p=5?

• Math100
In summary, the assertion that the only prime p for which 3p+1 is a perfect square is p=5 can be proven by supposing p is a prime and 3p+1=n^2 for some n##\in\mathbb{Z}##, and using the definition of a prime and the fundamental theorem of arithmetics to show that the only possible solution is p=5.f

#### Math100

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.

Delta2
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}##,
How is that? What about ##n=0,1,2##?
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.
You should use the definition of a prime here. We have ##3\,|\,3p\,|\,(n+1)(n-1) \Longrightarrow 3\,|\,(n-1) \text{ or }3\,|\,(n+1)## and go on from here.

Or you use the fundamental theorem of arithmetics which says that every number can be written uniquely as a product of primes. With that we get ##3=n-1## or ##3=n+1.## In both cases, you should discuss both cases, i.e. what about ##3=n+1?## If that is excluded, then you can go on with ##3=n-1.##

But you have used this theorem. With only using the definition of a prime, you can only conclude ##3\,|\,(n\pm 1).##

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Math100
How is that? What about ##n=0,1,2##?

You should use the definition of a prime here. We have ##3\,|\,3p\,|\,(n+1)(n-1) \Longrightarrow 3\,|\,(n-1) \text{ or }3\,|\,(n+1)## and go on from here.

Or you use the fundamental theorem of arithmetics which says that every number can be written uniquely as a product of primes. With that we get ##3=n-1## or ##3=n+1.## In both cases, you should discuss both cases, i.e. what about ##3=n+1?## If that is excluded, then you can go on with ##3=n-1.##

But you have used this theorem. With only using the definition of a prime, you can only conclude ##3\,|\,n\pm 1).##
Yes, I did a mistake, I didn't think about n=0, 1, 2. But is 0 a natural number? Because I thought that natural numbers start at 1 and above.

But is 0 a natural number?
Some say this, others say that. Personally, I do not consider it a natural number because I think it was a great achievement in human history when for the first time someone counted something that wasn't there. But I think most people nowadays start with ##0##. Anyway, since you haven't said what you mean by ##\mathbb{N}## both cases are possible.

Math100
Here's my revised proof for this assertion:

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).
Applying the Fundamental Theorem of Arithmetic produces:
n+1=3 or n-1=3.
Now we consider two cases.
Case #1: Suppose n+1=3.
Then we have n=3-1=2.
Thus 3p+1=n^2
=2^2
=4,
and so 3p=4-1=3,
which implies that p=3/3=1.
Case #2: Suppose n-1=3.
Then we have n=3+1=4.
Thus 3p+1=n^2
=4^2
=16,
and so 3p=16-1=15,
which implies that p=15/3=5.
Note that p=1 is not a prime number.
Therefore, the only prime p for which 3p+1 is a perfect square is p=5.

nuuskur, Delta2, FactChecker and 1 other person
Is it correct now? The above proof?

This looks good to me.

Math100
IMO, your proofs are certainly improving rapidly. The logic flow is good and you are providing the right amount of detail.

Math100, Delta2 and fresh_42
IMO, your proofs are certainly improving rapidly. The logic flow is good and you are providing the right amount of detail.
Thank you so much!

Thank you so much!
There are also some neat ideas appearing in your proofs. Perhaps the two things go together. Once you get the basics of logic sorted out you also start to get insights into specific problems.

Math100
There are also some neat ideas appearing in your proofs. Perhaps the two things go together. Once you get the basics of logic sorted out you also start to get insights into specific problems.
This is a good idea. Maybe this little problem is a nice challenge for you:

A Størmer number ##n## is a number for which there is a prime ##p## that is greater than ##2n## and a divisor of ##n^2+1##. E.g. ##n=16## and ##p=257.## We have ##257 > 2\cdot 16## and ##257\,|\,(16^2)+1=257##. Another one is ##n=33.## Can you figure out ##p## in this case?

Challenge: Prove that there cannot be a Størmer number of the form ##n=2k^2>2,## e.g. ##n=2\cdot 4^2=32## cannot be a Størmer number.

Another question: If we examine primes, what would be a natural distinction between two kinds of primes? And, no, odd and even doesn't get us very far.

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I'm confused, isn't ##k=1## going to give a Størmer number? I think it's the only one though.

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I'm confused, isn't ##k=1## going to give a Stromer number? I think it's the only one though.
##2\,|\,1^2+1## but ##2 \not\gt 2\cdot 1.##

##16## and ##33## are Størmer numbers.
Størmer numbers ##n## are exactly those numbers for which there isn't a linear combination
$$\operatorname{arccot}n =\sum_{k=1}^{n-1}a_k\cdot \operatorname{arccot}k\, , \,a_k\in \mathbb{Z}$$
Størmer numbers are therefore also called arc-cotangent irreducible numbers.

##2\,|\,1^2+1## but ##2 \not\gt 2\cdot 1.##
No, I meant ##2=2(1)^2## is a Størmer number since ##2^2+1=5, contrary to your post. I think it's the only one though.

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No, I meant ##2=2(1)^2## is a Stromer number since ##2^2+1=5, contrary to your post.
Where?
##2## is a Størmer number, so what? ##p\,|\,n## is not required.
I think it's the only one though.
What is wrong about ##16##?

I corrected the post. I had forgotten to demand ##n>2.##

This is a good idea. Maybe this little problem is a nice challenge for you:

A Størmer number ##n## is a number for which there is a prime ##p## that is greater than ##2n## and a divisor of ##n^2+1##. E.g. ##n=16## and ##p=257.## We have ##257 > 2\cdot 16## and ##257\,|\,(16^2)+1=257##. Another one is ##n=33.## Can you figure out ##p## in this case?

Challenge: Prove that there cannot be a Størmer number of the form ##n=2k^2>2,## e.g. ##n=2\cdot 4^2=32## cannot be a Størmer number.

Another question: If we examine primes, what would be a natural distinction between two kinds of primes? And, no, odd and even doesn't get us very far.
p in this case would be 109.

fresh_42 and SammyS