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.



Finding a solution for this system of two diophantine equations (depending on a parameter)
I think we can completely solve the problem. Since $\displaystyle\frac{3n+3x+n^2}{12}$ must be an integer, $y:=3x$ must be an integer. Put also $N:=n(n+3)$. Then we want


$$\frac{N+y}{12},\text{ and }\frac{N(N+y)}{12y}$$ to be positive integers.


Therefore we need $N+y=12r$ **(1)** and $y=\displaystyle\frac{N^2}{12k-N}$ **(2)**, with $r,k$ integers.


Now, we have $n=4m+1$, so $N=4(m+1)(4m+1)$, which is already a multiple of $4$. Due to **(1)**, $y$ must be also a multiple of $4$. Moreover, if:


* $m\equiv 2 (\text{mod} 3)$ then $N\equiv 0(\text{mod} 3)$, so we need $y\equiv 0(\text{mod} 3)$.
* $m\equiv 1 (\text{mod} 3)$ then $N\equiv 1(\text{mod} 3)$, so we need $y\equiv 2(\text{mod} 3)$.
* $m\equiv 0 (\text{mod} 3)$ then $N\equiv 1(\text{mod} 3)$, so we need $y\equiv 2(\text{mod} 3)$.


Let us find a valid value of $y$ from **(2)**. The first two cases are a bit simpler. We have:


$$y=4\frac{(m+1)^2(4m+1)^2}{3k-(m+1)(4m+1)}.$$


* $m\equiv 2 (\text{mod} 3)$: $m+1,4m+1$ are multiples of 3, hence for $y$ to be a multiple of 3 it is enough to leave one of the factors in the numerator (we choose the smallest one, $m+1$): pick $3k=(m+1)(4m+1)+(m+1)(4m+1)^2$ in order to get $y=4(m+1)$. This can be done since $(m+1)(4m+1)+(m+1)(4m+1)^2$ is a multiple of $3$.


We get $y=4m+4=n+3$ for $n=12t+9$.


* $m\equiv 1 (\text{mod} 3)$: $m+1,4m+1$ are congruent to 2 modulo 3, hence for $y$ to be 2 mod3 it is enough to leave one of the factors in the numerator (we choose the smallest one, $m+1$): pick $3k=(m+1)(4m+1)+(m+1)(4m+1)^2$ in order to get $y=4(m+1)$. This can be done since $(m+1)(4m+1)+(m+1)(4m+1)^2$ is a multiple of $3$.
Observe that this is not a happy coincidence: the numerator is congruent to 1 mod3, while the denominator is congruent to -1(mod 3)=2(mod 3), so that $y$ is congruent to $1/2=2$, as needed, and we just need to ensure that $k$ is such that the numerator is a multiple of the denominator.


We get $y=4m+4=n+3$ for $n=12t+5$.


* $m\equiv 0 (\text{mod} 3)$: This one is trickier. $m+1,4m+1$ are congruent to 1 modulo 3, and we need $y$ congruent to 2, so we have to get from the numerator a proper factor congruent to 2 (and then eliminate the rest using the denominator). If $m$ is odd, then we can factor out a 2 from the hanging 4 to get the desired congruence via $2(m+1)$ in the numerator, since then $m+1$ is even and $y$ is multiple of $4$. Thus:


If $m$ is odd we pick $3k=(m+1)(4m+1)+2(m+1)(4m+1)^2$ in order to get $y=2(m+1)=2(\frac{n-1}4 +1)$ for $n=24t+13$.


If $m$ is even (i.e.,multiple of 6), then there is a solution if and only if we can find a proper factor $F\equiv 2(\text{mod}3)$ inside $(m+1)^2(4m+1)^2$, iff $m+1$ or $4m+1$ contains such an $F$, in which case it contains another factor $G\equiv 2(\text{mod}3)$ (i.e., those factors come in pairs since $m+1,4m+1\equiv 1(\text{mod}3)$). Then $3k=(m+1)^2(4m+1)^2/F+(m+1)(4m+1)$ is valid and $y=4F$ is a solution (from this we infer that the possible solutions are of the form $y=12s+8$, $m+1$ or $4m+1$ multiple of $3s+2$).


For example, if:


* $m=6$ then $m+1=7$, $4m+1=25=5\cdot5$, $F=5$ does the job with $y=20$.
* $m=12$ then $m+1=13$, $4m+1=49=7\cdot7$ and there is no solution.
* $m=18$ then $m+1=19$, $4m+1=73$ and there is no solution.
* $m=24$ then $m+1=25=5\cdot 5$, $F=5$, $y=20$.


