# A squarefree integers conjecture

• Jul 20th 2010, 08:05 PM
Bacterius
[SOLVED] A squarefree integers conjecture
Hello,
I came up with a conjecture regarding squarefree integers of the form $\displaystyle x^2 + 1$. My conjecture is :

Quote:

$\displaystyle x^2 + 1$ is squarefree if and only if it is not a multiple of $\displaystyle 25$
Note : each prime in the prime factorization of a squarefree number only comes up once. For instance, $\displaystyle 2^2 \times 3$ is not squarefree while $\displaystyle 5^1 \times 11^1$ is.

I'm not absolutely sure it holds but I'm pretty confident, so I've been trying to prove it using an inductive step but I hardly see how to do it because as I try to do it (check its divisibility by the square of all prime numbers one after the other, and by some routine process show that it cannot be divided by it) it involves two variables which is obviously wrong ... perhaps someone here has an idea and could help me out on this one, give hints to get me started, put me on the right track (or you can just disprove the conjecture (Lipssealed)).
• Jul 20th 2010, 08:39 PM
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Quote:

Originally Posted by Bacterius
Hello,
I came up with a conjecture regarding squarefree integers of the form $\displaystyle x^2 + 1$. My conjecture is :

Note : each prime in the prime factorization of a squarefree number only comes up once. For instance, $\displaystyle 2^2 \times 3$ is not squarefree while $\displaystyle 5^1 \times 11^1$ is.

I'm not absolutely sure it holds but I'm pretty confident, so I've been trying to prove it using an inductive step but I hardly see how to do it because as I try to do it (check its divisibility by the square of all prime numbers one after the other, and by some routine process show that it cannot be divided by it) it involves two variables which is obviously wrong ... perhaps someone here has an idea and could help me out on this one, give hints to get me started, put me on the right track (or you can just disprove the conjecture (Lipssealed)).

Sorry to be bearer of bad news .. 239^2 + 1 = 2 * 13^4. My not-so-elegant method of arriving at that answer: going to Dario Alpern's Diophantine solver and trying (a,e,f) = (1, -4, 1), (1, -9, 1), (1, -16, 1), ... :( (Well -16 can be skipped since -4 produces no solutions.)
• Jul 20th 2010, 08:56 PM
Bacterius
Let's make an exception then xD
Joking, thanks Undefined, somehow I missed this one.
I'll put the topic as solved ... well not exactly but it's the only prefix we got yet.
• Jul 20th 2010, 10:42 PM
simplependulum
In fact , we can create infinitely many numbers that are not satisfying the condition .

By considering solutions of Pell's equation $\displaystyle x^2 - Ny^2 = -1$ , we know , for many $\displaystyle N$ , there are infinitely many solutions $\displaystyle (x,y)$ .

Therefore , $\displaystyle x^2 + 1 = Ny^2$ , $\displaystyle Ny^2$ is obviously not squarefree .

For example , $\displaystyle N = 2$ we have

$\displaystyle 7^2 + 1 = 2(5^2)$ Oh , which is included in your constraint but no problem , we have $\displaystyle (7+5\sqrt{2}) (3+2\sqrt{2}) = 41 + 29\sqrt{2}$ that means , $\displaystyle 41^2 + 1 = 2(29^2)$ .

There are many solutions for $\displaystyle x^2 - 2y^2 = - 1$ , all the solutions can be represented by this formula :

$\displaystyle x + y\sqrt{2} = (7+5\sqrt{2})( 3+2\sqrt{2} )^r$