$(n+1)(n+12)$ as a product of $4$ consecutive numbers

Question

The question is to find all natural numbers $n$, such that $(n+1)(n+12)$ can be written as a product of $4$ consecutive numbers.

My approach was the following: starting from the remark that if you add 1 to a product of four consecutive integers you receive a perfect square, I can write: $(n+1)(n+12)+1=k^2$, with $k \in \mathbb{N}$. Then, $n^2+13n+13=k^2 \iff 13(n+1)=k^2-n^2=(k-n)(k+n)$. Now, because $13$ is a prime, could I perhaps use the following logic: we might have two cases, either $k-n$ or $k+n$ should be 13(or 1?), and the other should be $n+1$. Solving for $n$ will give us $n=4$(I have however only considered the case where both $k-n, k+n$ are equal to 13, not 1. Is this approach correct, or could it be turned into a correct one?

Answer 1

Jjagmath indicated what is wrong with your approach. All that you have is $ 13 \mid k-n$ or $13 \mid k+n$.

If $ 13 \mid k-n$, then $n+1 \geq k+n$, which is a contradiction since $ k \geq \sqrt{ 2 \times 13 } > 5$. So $ 13 \mid k+n$.
However, I can't push through from here.

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Alternative solution: Observe that

$$ (n+3)^2 = n^2 + 6n + 9 < n^2 + 13 n + 13 < n^2 + 14n + 49 = (n+7)^2$$

Then, check $(n+j) ^2 = n^2 + 13n + 13 $ for $ j = 4$ to 6, so we just need to check these 3 cases.

• Checking $ j = 4$ yields $n = \frac{ 3}{5}$
• Checking $j = 5$ yields $n = 4$ .
  • This gives $k = j+n = 9$ and $ 13 \mid k+n = 13$.
  • $(n+1)(n+12) = 80$ is not a multiple of 3, hence cannot be written as the product of 4 consecutive integers.
• Checking $j = 6$ yields $n = 23$.
  • This gives $k = j+n = 29$ and $ 13 \mid k+n = 52$.
  • $(n+1)(n+12) = 4 \times 5 \times 6 \times 7$

Hence, the only solution is $n = 23$.

Note: We could tighten up the first inequality slightly. Since $ n \geq 1$, we also have $(n+4) = n^2 + 8n + 16 < n^2 + 13n + 13$, so we could reject $j = 4$.

Answer 2

One way continue your line of thought is as follows: If $(n+1)(n+12)$ is the product of four consecutive integers for some natural number $n$, then there is a natural number $k$ such that $$(n+1)(n+12)+1=k^2,\tag{1}$$ and because $n\geq0$ we have $k^2\geq13$ and so $k\geq4$. Rearranging $(1)$ shows that $$13(n+1)=k^2-n^2=(k+n)(k-n),$$ where $k+n>n+1$ and so $k-n<13$. In particular $13\mid k+n$, say $k+n=13m$. Then $$13(n+1)=(k+n)(k-n)=13m(13m-2n),$$ and again rearranging yields $$(2m+1)(26m-4n-13)=-9.$$ Of course $2m+1>0$ and so $2m+1\in\{1,3,9\}$, and $m\in\{0,1,4\}$ respectively. Then for the other factor we correspondingly find \begin{eqnarray} 26m-4n-13=-9\qquad&\text{ and }\qquad m=0\qquad&\text{ so }\qquad n=-1,\\ 26m-4n-13=-3\qquad&\text{ and }\qquad m=1\qquad&\text{ so }\qquad n=4,\\ 26m-4n-13=-1\qquad&\text{ and }\qquad m=4\qquad&\text{ so }\qquad n=23,\\ \end{eqnarray} Of course $n=-1$ is not a natural number, so this is not a solution. For $n=4$ we get $(n+1)(n+12)=5\times16=80$, but $$1\times2\times3\times4=24<80<120=2\times3\times4\times5,$$ so this is also not a solution. For $n=23$ we get $$(n+1)(n+12)=24\times35=2^3\times3\times5\times7=4\times5\times6\times7,$$ and so the unique solution is $n=23$.

Answer 3

First question: why $(n+1) \cdot (n+12)$? I would simply write this as $n \cdot (n+11)$ :-)

So, the question is: how to write $n \cdot (n+11)$ as $k \cdot (k+1) \cdot (k+2) \cdot (k+3)$.

I would go for three obvious possibilities:

1. $n = k \cdot (k+1)$ and $n+11 = (k+2) \cdot (k+3)$
2. $n = k \cdot (k+2)$ and $n+11 = (k+1) \cdot (k+3)$
3. $n = k \cdot (k+3)$ and $n+11 = (k+1) \cdot (k+2)$

The first equation becomes:
$(k+2) \cdot (k+3) - k \cdot (k+1) = 11$
$k^2+5k+6 - (k^2+k) = 11$
$4k+6 = 11$, which has no integer solutions.

The second equation becomes:
$(k+1) \cdot (k+3) - k \cdot (k+2) = 11$
$k^2+4k+3 - (k^2+2k) = 11$
$2k+3 = 11$, which has $k=4$ as an answer, leading to $n=24$ (or $n=23$ in your definition of $n$).
Indeed $(23+1) \cdot (23+12) = 4 \cdot 5 \cdot 6 \cdot 7$

The third equation becomes:
$(k+1) \cdot (k+2) - k \cdot (k+3) = 11$
$k^2 + 3k + 2 - (k^2 + 3k) = 11$
$2 = 11$, which also has no solutions :-)

Now, obviously: when the product of two first numbers equals the product of four second numbers, then it would be great if each of those two first numbers could be written as a product of two of those four numbers (which is the case I have elaborated), but this is not always the case, so my answer serves the purpose "Your question does have a solution." but not the purpose "Give me all the solutions.".