Finding a solution for this system of two diophantine equations (depending on a parameter) [closed]

Question

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.

Answer 1

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 $m+1=25=5\cdot 5$, $F=5$, $y=20$.

Answer 2

One can use common subexpressions to get a simple answer. Note that $a=n(n+3)$ must be a multiple of $4$. Setting $y=3x$, we look for $a+y$ is a multiple of $12$ and $a(a+y)$ is a multiple of $12y$. If we can pick $y$ to meet $a+y$ is a multiple of $12$, then $y$ is an integer and it suffices to also pick $y$ being a divisor of $a$. This is possible for $n=5$ and in general $y$ being a $2 mod 3$ divisor of $n(n+3)$ should work. I find $x=(n+3)/3$ works for many $n$.

Edit 2018.01.16

The problem is a little more intriguing. If $n$ is $0$ or $2 \equiv1 \pmod 3$, then letting $3x$ be $n+3$ gives a solution as can be easily checked. For $n=1 \equiv1 \pmod 3$ (and so $n$ is $1 \pmod {12}$, $3x$ needs to be an integer which is $\equiv8 \pmod {12}$ to satisfy the first relation, and setting $y=3x$ and $a=n(n+3)$ gives $12y$ has to divide $a(a+y)$. This is easy if $y$ divides $a$, while if $y$ does not divide a then with $d$ being the greatest divisor of $a$ and $y$, and $b =y/d$, we get $b$ has to divide $d$. For small $n$ with $n$ being $\equiv1 \pmod {12}$, we can find such a divisor $y$ which divides $a$, but it is not clear that we can always do that. So far $y=8$ or $20$ works for small $n$.

End Edit 2018.01.16

Gerhard "Addition More Complex Than Multiplication?" Paseman, 2018.01.16.

Answer 3

At first glance I thought that for $(1)$ to have solutions $n$ had to be a multiple of $3$, which is wrong when $x$ can be a rational, as Gerhard pointed out. So what follows is a solution when $n=3m$ for some $m$. Remain to be treated the cases $n\equiv1 \pmod 3$ and $n\equiv2 \pmod 3$.

So $n=3m$, with $m\equiv3 \pmod 4$.

Writing $m=4p+3$ condition $(1)$ becomes $$\dfrac{3(4p+3)+x+3(4p+3)^2}{4}\in \mathbb{N}$$

which is equivalent to $$\dfrac{1+x+3}{4}\in \mathbb{N}$$ so that $$(1)\Longleftrightarrow x\in 4\mathbb{N}$$.

Now if you plug in $n=3(4p+3)$ and $x=4q$ into $(2)$, you get massive simplifications into

$$\dfrac{(4p+3)(p+1)(3q+9(4p+3)(p+1))}{q}\in \mathbb{N}$$

showing that for instance any factor of $(4p+3)(p+1)$ will do.

Conclusion: when $n\equiv0 \pmod 3$

• $p=\dfrac{n-9}{12}$ is an integer,
• $x=4(p+1)$ is a solution.