what is the smallest $x,y,z,w$ where $x^2-1,y^2-1,z^2-1,w^2-1$ are pairwise coprime?

Question

Does there exist a set of four integers $x,y,z,w$ where $x^2-1,y^2-1,z^2-1,w^2-1$ are pairwise coprime? How could we expand this to $n$ variables ?

Answer 1

Others got you started. The existence of arbitrarily long such sequences can be proven without settling the twin prime conjecture - the old reliable Chinese remainder theorem will do the job instead.

Assume that you have found a sequence of $k$ integers $n_i,1\le i\le k$, such that the numbers $n_i^2-1$ are pairwise coprime.

Let $p_1,p_2,\ldots, p_m$ be all the prime factors of any of the numbers $n_i^2-1$, $i=1,2,\ldots,k$. We can select numbers $a_i$ such that $a_i\not\equiv \pm1\pmod{p_i}$. This is obvious when $p_i>2$ (because only two residue classes are forbidden), and $a_i=0$ works when $p_i=2$.

By the Chinese remainder theorem there exists an integer $n$ such that $$ n\equiv a_i\pmod{p_i} $$ for all $i=1,2,\ldots,m$. This implies that neither $n+1$ nor $n-1$ is divisible by any of the primes $p_i,i=1,2,\ldots,m$. Therefore selecting $n_{k+1}=n$ allows us to continue with the sequence.

Answer 2

$2^2-1$, $6^2-1$, $12^2-1$, $18^2-1$.

It seems (Twin prime problem) that there are infinitely many these numbers.

Answer 3

Consider the pairs of twin primes: $(3,5), (11,13), (17,19),(29,31)$. Then set $x = 4, y = 12, z = 18, w = 30$. Then $x^2 - 1, y^2 - 1, z^2 - 1, w^2 - 1$ because each has two prime factors and all the prime factors are different.

To be pedantic: I don't know if this is the smallest example. However you only asked for the existence of $(x,y,z,w)$ in your question. So by the Well Ordering Principle, since I found one example, there exists a smallest example.

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This also works for $n$ integers $x_1, \cdots, x_n$. Just find $n$ pairs of twin primes $(p_i, p_i + 2)$ (where each prime is distinct), and set $x_i = p_i + 1$. However since the twin prime conjecture hasn't been proven, you can't be sure that this works for arbitrarily large $n$.