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As an amateur I am not quite sure should I post a question on the site for professional mathematicians but if the question is not appropriate for this site you can freely migrate it to MathematicsStackExchange.

I was thinking about (integer) Diophantine $m$-tuples, and (integer) Diophantine $m$-tuple is a set of $m$ positive integers $\{a_1,a_2...,a_m\}$ such that $a_i \cdot a_j +1$ is a perfect square for all $1\leq i <j \leq m$.

It was known for a long time that there exist an infinite number of (integer) Diophantine $4$-tuples and it was conjectured that there does not exist a (integer) Diophantine quintuple but in this paper it is claimed that the problem of quintuples is solved and on this page you can find a lot of information about Diophantine $m$-tuples and various generalizations.

The generalization that I created while I was thinking about (ordinary/classical) Diophantine $m$-tuples is this one:

For every $(k,l) \in \mathbb Z^{2}$ we can look at the (integer) Diophantine problem of $m$-tuples such that $a_i \cdot a_j +ka_i + la_j +1$ is a perfect square for all $1\leq i <j \leq m$ and we see that the classical Diophantine $m$-tuples are just the special case that corresponds to $(k,l)=(0,0)$. Of course that this generalization can be further generalized but at the moment I do not see the need to talk about a generalizations that are not at the heart of the question that I am going to ask and the question is:

Is it true that for every natural number $c \geq2$ there exists at least one pair $(k_c,l_c) \in \mathbb Z^{2}$ and a set of $c$ positive integers $\{a_1,a_2,...,a_c\}$ such that $a_i \cdot a_j +k_ca_i + l_ca_j +1$ is a perfect square for all $1\leq i <j \leq c$?

Is this conjecture true?

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  • $\begingroup$ No. You can reduce the number of equations, but the task will be reduced to systems of nonlinear equations. And there is the solvability if the number of equations in integers is not possible. Each specific system has emerged solvability for a long time. Therefore it is better to start with a small number of equations and to increase them. $\endgroup$ – individ May 1 '17 at 5:20
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The conjecture is trivial -- just take $k_c=l_c=1$ and $a_i = (i+1)^2-1$ for all $i=1,2,\dots,c$. Then $$a_i \cdot a_j +k_ca_i + l_ca_j +1 = ((i+1)(j+1))^2.$$

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