MHB How Many Distinct Numbers Are in This Mathematical Sequence?

[MarkFL]

anemone has asked me to fill in for her this week.

Here is this week's POTW:

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How many distinct numbers are in the list

$$\frac{1^2-1+4}{1^2+1},\,\frac{2^2-2+4}{2^2+1},\,\frac{3^2-3+4}{3^2+1},\,\cdots,\frac{2011^2-2011+4}{2011^2+1}$$

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[MarkFL]

Congratulations to the following members for their correct solutions:

• kaliprasad
• lfdahl

Here is the solution anemone provided:

Spoiler

We can rewrite the general term of this sequence that contains for a total of $$2011$$ terms as

$$a_n=\frac{n^2-n+4}{n^2+1}=1-\frac{n-3}{n^2+1}$$, where $n\in \mathbb{Z}^+.$

We want to look for the terms, says, $$a_n$$ and $$a_m$$ that have the same value and see how many such pairs are there in the given sequence.

Since $$a_n$$ and $$a_m$$ are equal, we get:

$$1-\frac{n-3}{n^2+1}=1-\frac{m-3}{m^2+1}$$

Upon simplifying gives:

$$\frac{n-3}{n^2+1}=\frac{m-3}{m^2+1}$$

$$(n-3)(m^2+1)=(m-3)(n^2+1)$$

$$nm^2+n-3m^2-3=mn^2+m-3n^2-3$$

$$nm^2-mn^2-m+n-3m^2+3n^2=0$$

$$nm(m-n)-(m-n)-3(m^2-n^2)=0$$

$$nm(m-n)-(m-n)-3(m-n)(m+n)=0$$

$$(m-n)(nm-1-3(m+n))=0$$

We know $n=m$ leads to the trivial case, we will assume that $n\ne m$.

Hence we get $nm-1-3(m+n)=0\implies m(n-3)=3n+1$. From here we can tell $n\ne 3$ or it leads us to $3n+1=0$, which is a contradiction.

Thus, $$m=\frac{3n+1}{n-3}=3+\frac{10}{n-3}$$. Since $m$ is an integer, we know the denominator $n-3$ has to divide $10$, we have only four possibilities:

$n-3=1,\,\implies n=4,\,m=13$

$n-3=2,\,\implies n=5,\,m=8$

$n-3=5,\,\implies n=8,\,m=5$

$n-3=10,\,\implies n=13,\,m=4$

We therefore find there are only two pairs of terms that are equal, namely $a_4=a_{13}=\dfrac{16}{17}$ and $a_5=a_{8}=\dfrac{12}{13}$.

We can conclude by now that there are $2011-2=2009$ distinct numbers in the given sequence.