# $x^2-1$ with prime factors $< 100$

I am reading at a paper about all integers $x$ such that $x^2-1$ has only prime factors smaller than $100$ (Luca and Najman).

My problem is that it is not obvious to me that the number of $x$ would be finite. I get the feeling the authors think it is too simple to be explained, but I still wonder if someone might do me a favor of demonstrating it to me.

• I think this is this paper: math.leidenuniv.nl/~fnajman/FLFNMC.pdf – miracle173 Feb 27 '14 at 17:49
• @miracle173, you are right that the paper explains why there is a finite number only. And that is presumably the paper OP is referencing. – vonbrand Feb 27 '14 at 17:59
• You are probably both right, it is in any case the article I am referring to. My problem is probably just that, as an amateur, I need some extra steps in the proof. – Mikael Jensen Feb 28 '14 at 10:02

I gave the issue some additional thought, using Pell’s equation for fixed d in the equations below (you might want to check the proof):

Using Brahmagupta’s technique, we can combine two Pell’s equations $((x_1)^2-d(y_1)^2)=1$ and $((x_2)^2-d(y_2)^2)=1$ to get $((x_1)^2-d(y_1)^2)((x_2)^2-d(y_2)^2)=((x_1)(x_2)+d(y_1)(y_2))^2 - d((x_1)(y_2+(x_2)(y_1))^2$

The second lowest solution whose x component we now call $x_2$ can be achieved by combining the first solution with itself to get the first higher solution $(x_2,y_2,d)$ and get $x_2=2x_1^2-1$ (using the substitution $d(y_1)^2)= (x_1)^2-1$, and $y_2=2(x_1)(y_1)$.

We observe that $(x_2)^2-1=(2x_1^2-1)^2-1=$ and $4(x_1)^2 ((x_1)^2-1) =d(y_2)^2=d(2(x_1)(y_1))^2$.

The LHS in the last equation contains the factor $(x_1)^2-1$. This implies that this factor ($(x_1)^2-1$ ) does not contain any of the primes making up $x_1$ since $(x_1)^2-1=-1$ (mod p) for all the primes p in x.

If $x_2$ is combined with $x_1$ we get the third lowest solution $x_3$: $(x_3)^2-1= d(y_3)^2= (x_3)^2-1= ((x_1)*(x_2)+d*(y_1)*(y_2))^2-1=((x_1)*(2*(x_1)^2-1)+d*(y_1)*(2(x_1)*(y_1)))^2-1$

so that $(x_3)^2-1$ equals a constant $C(x_1)-1$. As above, this shows that all successive values of x for the chosen d must contain new primes. Since d also is subject to the same limitation being part of $(x_3)^2-1$ there must be a limited number of x, such that $x^2-1$ contains a limited number of primes. (I have used– without proof – the fact that the combination process yields all possible solutions of Pell’s equation for a given d. I realise it may have been more elegant using $x ± \sqrt{d}y$).

Edit: I now see that the second part of the proof only tells us that higher combinations are different from the nearest lower - the fact that y's highest prime factor increases for increasingly higher combinations (as they do), does not follow from the proof, as I first thought.

• Perhaps you could use x_1 and \sqrt{d} to improve the readability. – Bart Michels Mar 7 '14 at 16:30
• Thank you. I would be fully educated - since I only use modest equations - if I also knew how to write x^(n+1) the right way. The dollar sign will not do the trick. – Mikael Jensen Apr 1 '14 at 20:51
• You can use curly braces like {} to group things. For example, x^{n+1} or x_{n+1}. – Bart Michels Apr 2 '14 at 14:51
• Thank you, this completes my education but I just have to test: – Mikael Jensen Apr 4 '14 at 21:46
• Thanks.I just have to test: $1^{2+3}$ - I've got it! – Mikael Jensen Apr 4 '14 at 21:48