Multiples that are one less than Squares

Question

I was inspired to ask this problem after trying to find all $(x,y,u,v)$ for which $xy+1,xu+1,xv+1,yu+1,yv+1,uv+1$ are all sqaure.

After some basic calculation I was easily able to find $x=n, y=n+2, u=4n+4$. However, I had some difficulty finding $v$.

It took some time, but I was able to find $v=4(n+1)(2n+1)(2n+3)$.

Are there more easier ways of finding $(x,y,u,v)$?

And is the following generlization possible: There exists infinitely many sets $X_{ n }=\{ x_{ i }|1\le i\le n \quad i\in \mathbb{N}\}$ and ${ X }_{ n }\subset \mathbb{N}$ for any finite natural number $n$ for which ${ x }_{ i }x_{ j }+1$ are all perfect squares?

Answer 1

This problem is well-known in the literature, where such sets are called Diophantine $4$-tuples. Dujella has made an extensive study of them: see his web page at https://web.math.pmf.unizg.hr/~duje/dtuples.html for plenty of references.

In particular, Dujella proved that there are no Diophantine $6$-tuples and that any $5$-tuples, if they exist, are bounded by an absolute constant so there are at most finitely many (none have been found). So the conjecture in your last paragraph is false for every $n \ge 6$.

Your question about extending a triple of integers to a $4$-tuple has also been considered (see https://web.math.pmf.unizg.hr/~duje/quint.html for the reference). We may always take

$$v = x + y + z + 2xyz + 2\sqrt{(xy+1)(yz+1)(xz+1)}.$$

It's conjectured that the above value of $v$ is unique (which would immediately imply that $4$ is the largest possible Diophantine tuple).

Answer 2

It is better to use a more General approach. We write the system.

$$\left\{\begin{aligned}&xy+T=a^2\\&xz+T=b^2\\&xq+T=c^2\\&yz+T=d^2\\&yq+T=k^2\\&zq+T=n^2\end{aligned}\right.$$

If the number $T$, lay at the multipliers. We find then the desired settings.

$$T=3(p-t-s)(p+t-s)(p+s)^2$$

Then the solution can be written as.

$$x=t^2+2s^2+2ps-p^2$$

$$y=t^2-s^2+2ps+2p^2$$

$$z=4t^2-(p-s)^2$$

$$q=3(p+s)^2$$

$$a=p^2+ps+s^2-t^2$$

$$b=2t^2+s^2+ps-2p^2$$

$$c=3(p+s)s$$

$$d=2t^2-2s^2+ps+p^2$$

$$k=3(p+s)p$$

$$n=3(p+s)t$$

Answer 3

I thought as this task to generalize and use for any numbers. It turned out that you can do without calculations. For the system of equations:

$$\left\{\begin{aligned}&ab+T=x^2\\&ac+T=y^2\\&bc+T=z^2\end{aligned}\right.$$

Enough to factor the following number:

$$bc=(y+c)^2-T$$

Using these numbers you can easily write the solution of this system of equations.

$$a=b-c-2y$$

$$b=b$$

$$c=c$$

$$x=b-c-y$$

$$y=y$$

$$z=y+c$$