I am trying to find $a$ such that $a^2+1$ is not squarefree, meaning that none of its factors are square numbers. I found that this is the case for $a=7$ but can't think of a way to approach this problem more generally.
Asked
Active
Viewed 98 times
0
-
1What do you mean by squarefree? – user Jan 25 '18 at 21:58
-
@gimusi not divisible by the square of an integer – eyeballfrog Jan 25 '18 at 21:59
-
1You want to pick a prime $p$ such that $-1$ is a quadratic residue mod $p$. These are $p=2$ and $p\equiv 1\mod{4}$. Then, for those $p$ solve the congruence $a^2+1\equiv0\mod{p^2}$. – orole Jan 25 '18 at 21:59
-
@gimusi Squarefree means that it has no repeated prime factors. – orlp Jan 25 '18 at 21:59
-
Are you trying to find all such $a$, or an easy construction that always generates them? – orlp Jan 25 '18 at 22:00
-
Do you mean the prime decomposition of $a^2 + 1$ is not square free? It looks like you have found such an a in which $a = 7 \to a^2 + 1 = 50 = 2\cdot 5^2$. You have found such an $a$. If you're asked to find a counterexample, to show that $a^2 + 1$ is not always square free, you've found it. – amWhy Jan 25 '18 at 22:01
-
Possible duplicate of Infinitely many $n$ such that $n^2+1$ is squarefree – QuIcKmAtHs Jan 25 '18 at 22:02
-
Check out Pell's equation to generate lots more examples ... Eg $29^2 \mid 41^2+1$. – Donald Splutterwit Jan 25 '18 at 22:04
-
@XcoderX This question is looking for $a^2 + 1$ that is NOT square free, and the asker found one such $a$, as asked. – amWhy Jan 25 '18 at 22:04
-
1It is similar... – QuIcKmAtHs Jan 25 '18 at 22:04
-
@orlp I'm trying to find all such $a$. – user524960 Jan 25 '18 at 22:08
1 Answers
2
For any prime $p\equiv 1\pmod{4}$ you have that $a^2+1\equiv 0\pmod{p}$ has two opposite solutions $\pmod{p}$, and they can be lifted to solutions $\!\!\pmod{p^2}$. For instance, by picking $p=13$ we have that the square roots of $-1$ in $\mathbb{F}_p$ are $5$ and $8$. Let us see if for some $k\in\mathbb{Z}$
$$ (13k+5)^2\equiv -1\pmod{13^2} $$
holds. Of course, since the previous equation is equivalent to
$$ 10k+2 \equiv 0\pmod{13} $$
which is solved by $k\equiv 5\pmod{13}$.
It follows that $(5\cdot 13+5)^2+1=70^2+1$ is a multiple of $13^2$.
Jack D'Aurizio
- 353,855
-
Thank you. Does this generate all cases? I mean, are there any cases with non-prime square factors? – user524960 Jan 25 '18 at 22:12
-
1@user524960: all the non-squarefree numbers of the form $a^2+1$ can be generated by this procedure, since the odd prime divisors of $a^2+1$ are always of the $4k+1$ form. – Jack D'Aurizio Jan 25 '18 at 22:15