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Post Closed as "Not suitable for this site" by Henry.L, David Handelman, Max Alekseyev, Alexey Ustinov, Mikhail Katz

About Finding a common pointsolution for positive integersthis system of two diophantine equations (depending on a parameter)

Added top level nt.number-theory tag
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Joe Silverman
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I propose the following problem (Maybe it has a trivial solution):

Let $n$ be a positive integer such that $$n=1 \mod 4$$$$n\equiv1 \pmod 4.$$

Then the problem is about findingto find a rational $x$ inas a function of $n$ such that $ \dfrac{3n+3x+n^{2}}{12}$ and $\dfrac{n(n+3)(3n+3x+n^{2})}{36x}$ are $$ \dfrac{3n+3x+n^{2}}{12} \quad\text{and}\quad \dfrac{n(n+3)(3n+3x+n^{2})}{36x}$$ are both strictelystrictly positive integers. Or at least how one can proves that such a $x$ exists.

I propose the following problem (Maybe it has a trivial solution):

Let $n$ be a positive integer such that $$n=1 \mod 4$$

Then the problem is about finding rational $x$ in function of $n$ such that $ \dfrac{3n+3x+n^{2}}{12}$ and $\dfrac{n(n+3)(3n+3x+n^{2})}{36x}$ are both strictely positive integers. Or at least how one can proves that such a $x$ exists.

I propose the following problem (Maybe it has a trivial solution):

Let $n$ be a positive integer such that $$n\equiv1 \pmod 4.$$

Then the problem is to find a rational $x$ as a function of $n$ such that $$ \dfrac{3n+3x+n^{2}}{12} \quad\text{and}\quad \dfrac{n(n+3)(3n+3x+n^{2})}{36x}$$ are both strictly positive integers. Or at least how one can proves that such a $x$ exists.

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Safwane
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About a common point for positive integers

I propose the following problem (Maybe it has a trivial solution):

Let $n$ be a positive integer such that $$n=1 \mod 4$$

Then the problem is about finding rational $x$ in function of $n$ such that $ \dfrac{3n+3x+n^{2}}{12}$ and $\dfrac{n(n+3)(3n+3x+n^{2})}{36x}$ are both strictely positive integers. Or at least how one can proves that such a $x$ exists.