Three quarters of MOP got 7, making this the easiest problem on TSTST.

(a) We first show that $n\in S$ iff there are integers $w,x\geq 2$ such that $n=w(w-1)x(x-1).$ Note that if $p(x)=x^2+mx+n$ has an integer root, its other root is necessarily an integer by Vieta's. Moreover, both roots are negative, so if $m,m+1\in \mathcal{F}(n),$ we may let \[x^2+mx+n=(x+a)(x+b),x^2+(m+1)x+n=(x+c)(x+d),a,b,c,d\in \mathbb{N}.\]This implies $ab=cd=n,$ so by the Factor Lemma there are positive integers $w,x,y,z$ satisfying $a=wx,b=yz,c=wy,d=xz.$ Then \[a+b+1=c+d\iff wx+yz-wy-xz=1\iff (w-z)(x-y)=1\implies z=w-1,y=x-1\](or they're both -1, but then we can just rename the variables). This proves the characterization. Now all $n\in S$ has at least one such representation, so \[\sum_{n\in S}\frac{1}{n}\leq\sum_{w,x\geq 2}\frac{1}{w(w-1)x(x-1)}=\left(\sum_{x\geq 2}\frac{1}{x(x-1)}\right)^2=\left(\sum_{x\geq 2}\frac{1}{x-1}-\frac{1}{x}\right)^2=1,\]as desired. b) We claim that $\mathcal{F}(n)$ for $n$ of the form $6s(s+1)(2s+1)(3s+2),s\in \mathbb{N}$ have three consecutive integers, which will prove the statement. Indeed, note that \[(x+6s(s+1))(x+(2s+1)(3s+2))=x^2+(12s^2+13s+2)x+n\]\[(x+2s(3s+2))(x+3(s+1)(2s+1))=x^2+(12s^2+13s+3)x+n\]\[(x+3s(2s+1))(x+2(s+1)(3s+2))=x^2+(12s^2+13s+4)x+n\]imply that $12s^2+13s+\{2,3,4\}\in \mathcal{F}(n).$

Another construction for b): $n=k(k+1)^2(k+2)^2(k+3)$ satisfies the conditions since \begin{align*} (x+k(k+2)(k+3))(x+(k+1)^2(k+2)) &= x^2+(2k^3+9k^2+11k+2)x+n \\ (x+(k+1)^2(k+3))(x+k(k+2)^2) &= x^2+(2k^3+9k^2+11k+3)x+n \\ (x+(k+1)(k+2)^2)(x+k(k+1)(k+3)) &= x^2+(2k^3+9k^2+11k+4)x+n \\ \end{align*}

A different construction (found using Pythagorean triplets) for b): Let $k \in \mathbb{Z^+}$ and $n =k^2(k^2-1)(4k^2-1)$. Then $m =2k(2k^2-1)$ satisfies the given condition.

For (a) it is easy to show that $n\in S $ iff $n=a (a+1)d (d+1) $ the rest is too simple using $\frac {1}{a (a+1)}=\frac {1}{a}-\frac{1}{a+1} $

NOTE:The whole problem is just Four Numbers Theorem

My construction for (b): By Pell's there exists infinitely many positive integer solutions $(a,b)$ to $a^2-2b^2=7$. Note that $b$ must be odd. Let $y=\frac{b+1}{2}$ and $x=a-3y+2$, both are integers. Note that we can consider only large enough solutions $(a,b)$ so that $y$ and $-x$ are arbitrary large (and positive.) Then there will exist integers $r,s$ that $r+s=2xy-x-y+2$ and $rs=x(x-1)y(y-1)$, done.

Fun question: Does there exists positive integer $N$ such that $\mathcal{F} (n)$ contains no consecutive $N$ elements for any $n\in \mathbb{Z}^+$?

If $p$ has integer roots, then it can be factored as $\left(x+a\right)\left(x+\frac{n}{a}\right)$ where $a \mid n$. Then $m = a + \frac{n}{a}$, so if $\mathcal F(n)$ contains two consecutive integers, then there exists two distinct integers $a$, $b$ such that $a \mid n$, $b \mid n$, and \[a + \frac{n}{a} - b - \frac{n}{b} = 1.\]Then $n = ab+ \tfrac{ab}{b-a}$ so if $a \mid \tfrac{ab}{b-a}$, then $b-a \mid b$. Then if $x = b-a$, let $x y = b$ for some integer $y$ so that $n = (xy-x)(xy) + (xy-x)(y) = x(x-1)y(y-1)$ for $x,y \in \mathbb{Z}$. It's easy to see that all $n$ in this form satisfy the relation so $S$ is infinite. Since all $n$ is in the form $x(x-1)y(y-1)$, we have \[ \sum_{n \in S} \frac{1}{n} = \sum_{x,y >1} \frac{1}{x(x-1)y(y-1)} = \left(\sum_{x>1}^{ }\frac{1}{x-1}-\frac{1}{x}\right)\left(\sum_{y>1}^{ }\frac{1}{y-1}-\frac{1}{y}\right) = 1 \]which completes part a. For part $b$, I claim that all $n$ in the form $(2t+1)(2t)(3t+3)(3t+2)$, $t \in \mathbb{N}$ work. We have \[(2t)(3t+3)+(2t+1)(3t+2) = 12t^2+13t+2 \]\[(2t)(3t+2)+(2t+1)(3t+3) = 12t^2+13t+3\]\[(3t)(2t+1) + (2t+2)(3t+2) = 12t^2+13t+4\]as desired. $\blacksquare$

(a) Suppose both $x^2+mx+n$ and $x^2+(m+1)x+n$, so $m,m+1\in\mathcal{F}(n)$. Then, suppose the first had integer root $-a$, and second had $-b$. Therefore, $a+n/a=m$ and $b+n/b=m+1$, so \[b+\frac{n}{b}=a+\frac{n}{a}+1\implies n(1/a-1/b)=b-a+1\implies n=ab\left(1+\frac{1}{b-a}\right).\]We need $n/a\in\mathbb{Z}$, so $b/b-a\in\mathbb{Z}$, so suppose $b=k(b-a)$. Therefore, $b(k-1)=ka$, so $b=\ell k$ and $a=\ell(k-1)$ for some $\ell\in\mathbb{Z}$. Therefore, \[n=\ell^2k(k-1)(1+1/\ell)=\ell(\ell+1)k(k-1).\]Thus, $n=x(x+1)y(y+1)$ for positive integers $x,y$. It is easy to see then that $xy+(x+1)(y+1),x(y+1)+y(x+1)\in\mathcal{F}(n)$, so this works for all $x,y$. We now note that \[\sum_{n\in S}\frac{1}{n}\le\sum_{x\ge 1}\sum_{y\ge 1}\frac{1}{x(x+1)}=\left(\sum_{x\ge 1}\frac{1}{x(x+1)}\right)^2=1,\]as desired. (b) I have a really bad construction, but it works. The motivation was to get $m=2xy+x+y-1$ to work as well, then bashing out $m^2-4n$ is a square, difference of squares, and then factor. Note that $m\in\mathcal{F}(n)$ if and only if $m^2-4n$ is a perfect square (this can be seen from the quadratic formula). Let $n=x(x+1)y(y+1)$ where $y=4+x(x+1)/2+3x+1$. Obviously $2xy+x+y,2xy+x+y+1\in\mathcal{F}(n)$. We claim that $2xy+x+y-1\in\mathcal{F}(n)$ as well. It is a straightforward computation to verify that \[(2xy+x+y-1)^2 - 4x(x+1)y(y+1) = \left(\frac{x(x+1)}{2}-4\right)^2\]where $y=x(x+1)/2+4+3x+1$, and then we are done as the RHS is a square. This can be done by hand, but here is a verification: https://www.wolframalpha.com/input/?i=simplify+(2xy%2Bx%2By-1)%5E2+-+4x(x%2B1)y(y%2B1)+-+(x(x%2B1)%2F2-4)%5E2+where+y%3Dx(x%2B1)%2F2%2B4%2B3x%2B1

My solution to (a) is coincide with yayups'. But here is even worse construction for part (b) . Fix $k\in\mathbb{N}$ and $n = 8k^2(k-1)(2k-1)^2(2k+1)$. Consider \begin{align*} (X-4k(k-1)(2k+1))(X-2k(2k-1)^2) &= X^2 - (16k^3-12k^2-2k)X + n \\ (X-(2k-1)^2(2k+1))(X-8k^2(k-1)) &= X^2 - (16k^3-12k^2-2k+1)X + n \\ (X-4k^2(2k-1))(X-2(k-1)(2k-1)(2k+1)) &= X^2 - (16k^3-12k^2-2k+2)X + n \end{align*}so we are done.

Yet another construction for part $(b):$ If $n=x(x+1)^2(x+2)(2x+1)(2x+3),$ we claim that $\mathcal{F}(n)$ has three consecutive elements. Notice that for any $t|a,$ we have $t+\tfrac{n}{t} \in \mathcal{F}.$ So the elements $$(x+1)^2(2x+1)+x(x+2)(2x+3)=4x^3+12x^2+10x+1$$$$x(x+1)(2x+3)+(x+1)(x+2)(2x+1)=4x^3+12x^2+10x+2$$$$x(x+2)(2x+1)+(x+1)^2(2x+3)=4x^3+12x^2+10x+3$$are all consecutive terms in $\mathcal{F}(n),$ so we done. $\square$

how.

surprised that this was a tstst problem; not bad, wouldve made a beautiful jmo problem

Here's my attempt at remembering my awful in-contest solution. Who needs NT? There are infinitely many by considering $n=k^2(k+1)^2$ since $k^2+(k+1)^2=k(k+1)+k(k+1)+1$. However, if $a<b\leq \sqrt{n}$ satisfy $$\frac{n}{a}+a=\frac{n}{b}+b+1$$then we can get $n=\frac{ab(1+(b-a))}{b-a}$, and $$n \geq b^2 \Rightarrow \frac{ab(1+b-a)}{(b-a)}\geq b^2 \Rightarrow (b-a)^2+(b-a)a \leq a+(b-a)a \Rightarrow (b-a)^2 \leq a$$so the sum is at most $$\sum_{a=1}^{\infty}\sum_{x\leq \sqrt{a}} \frac{x}{(1+x)a(a+x)}=\sum_{x=1}^{\infty}\frac{1}{x+1} \sum_{a=x^2}^{\infty} \frac{1}{a}-\frac{1}{a+x}$$$$=\sum_{x=1}^{\infty} \frac{1}{x+1}\left(\frac{1}{x^2}+\frac{1}{x^2+1}\ldots+\frac{1}{x^2+x-1}\right) \leq \sum_{x=1}^{\infty} \frac{1}{x+1}\frac{x}{x^2}$$which telescopes to one by standard techniques, as desired. I dont remember it being this clean in contest so maybe I did something worse. Then for (b) i think i just did pell like everyone else

Strange problem. (a) We attempt to characterize $S$. Let $rs=n,r+s=m$ and $r's'=n,r'+s'=m+1$. Assume $r'<r$. Write \[r+\frac{n}{r}=m,r'+\frac{n}{r'}=m+1\implies (r'-r)+n(r-r')/rr'=1.\]The latter rearranges to \[1-(r'-r)=n(r-r')/rr'\implies n = \frac{rr'+r^2r'-rr'^2}{r-r'}.\]In particular, setting $r=r'+1$ yields an infinite family of solutions of the form \[n=r(r-1)+r^2(r-1)-r(r-1)^2 = r^2-r+r^3-r^2-r^3+2r^2-r=2r^2-2r.\]Let $r-r'=k$, then write \[n=rr'\cdot\frac{k+1}{k}=r'(r'+k)(1+1/k).\]Note that we must have $r\mid n,r'\mid n$, so $\frac{r'}{k}$ must be an integer. Then, we have the bound \[\sum_{n\in S}\frac{1}{n}\le \sum_{r'=1}^\infty \sum_{k\mid r'} \frac{k}{k+1}\cdot\frac{1}{r'(r'+k)}=\sum_{k=1}^\infty \frac{1}{k+1}\sum_{k\mid r'} \frac{k}{r'(r'+k)}=\sum_{k=1}^\infty \frac{1}{k+1}\sum_{k\mid r'} \left(\frac{1}{r'}-\frac{1}{r'+k}\right)=\]\[\sum_{k=1}^\infty \frac{1}{k+1}\cdot\frac{1}{k} = \sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)=1,\]as desired. (b) Now, in the characterization suppose that there are two possible values of $r$; $r=r'+k,r'-\ell$. Then, write the equality \[n=r'(r'+k)(1+1/k)=r'(r'-\ell)(1+1/\ell).\]Then, expand \[r'(1+1/k)+k+1=(r'+k)(1+1/k)=(r'-\ell)(1+1/\ell)=r'(1+1/\ell)-\ell-1.\]Rearranging, we have \[k+\ell+2=r'(1/\ell-1/k)=(k-\ell)\frac{r'}{k\ell}.\]Take $k=\ell+1$ to yield the curve of solutions $r'=\ell(\ell+1)(2\ell+3)$ with \[n=\ell(\ell+1)(2\ell+3)(\ell(\ell+1)(2\ell+3)-\ell)(1+1/\ell)=\ell(\ell+1)^2(2\ell+3)(2\ell^2+5\ell+2).\]

a) By looking at the discriminant we want there to be some $m$ such that there are $k,\ell$ with $$(m+k)(m-k)=4n=(m+1+\ell)(m+1-\ell) \Rightarrow (\frac{m+k}{2})(\frac{m-k}{2})=n=(\frac{m+1+\ell}{2})(\frac{m+1-\ell}{2})$$i.e. we want to factor $n$ into two pairs of factors whose sums differ by $1$. Let the smallet number in these two pairs be $t$, and let the pairs be $(t,t+a+b+1),(t+a,t+b)$. Since these two pairs both multiply to $n$ we get $t=ab \Rightarrow n=ab(a+1)(b+1)$. All such $n$ indeed work (with $m=2ab+a+b$), and the sum is $$\displaystyle\sum\limits_{a \geq 1} \displaystyle\sum\limits_{b \geq 0} \frac{1}{ab(a+1)(b+1)} =\displaystyle\sum\limits_{a \geq 1} \left( \frac{1}{a^2+a} \displaystyle\sum\limits_{b \geq 1} \frac{1}{b}-\frac{1}{b-1} \right) = 1$$ b) When $n=6t(t+1)(3t+2)(2t+1)$ the numbers $m=12t^2+13t+2, 12t^2+13t+3, 12t^2+13t+4$ work.

I have a different way to solve part (b). For some positive integers $x$ and $y$, let $ab = (a-x)(b+x+1) = (a-y)(b+y+2)$, so $$a(x+1) = bx + x(x+1)$$and $$a(y+2) = by+y(y+2).$$Multiplying the first equation by $y+2$ and the second by $x+1$ gives $$a(x+1)(y+2) = bx(y+2)+x(x+1)(y+2) = by(x+1)+y(y+2)(x+1)$$$$\implies b(2x-y) = (x+1)(y+2)(y-x).$$We want to prove that there are infinite many $x$ and $y$ such that there exists a positive integer solution for $a$ and $b$. If we let $2x-y = y-x$, we get $3x = 2y$, and substituting this gives $b = (x+1)(y+2)$, which means $a = x(y+3)$, and we are done.

Worked with dantaxyz. Claim: We can fully classify $S$ as follows: \[ S=\{x(x+1)y(y+1):x,y\in \mathbb{Z}_{>0}\}.\]Proof: $(\impliedby)$ Given $n=x(x+1)y(y+1)$, choosing $m=2xy+x+y$ works: \begin{align*} m&= x(y+1)+y(x+1) \\ m+1 &= xy+(x+1)(y+1). \end{align*}$(\implies)$ If $n\in S$, then there exists some $m$ such that $x^2+mx+n=(x+a)(x+\tfrac{n}{a})$ and $x^2+(m+1)x+n=(x+b)(x+\tfrac{n}{b})$ for some $a,b \in \mathbb{Z}$. So $a+\tfrac{n}{a}=m$ and $b+\tfrac{n}{b}=m+1$. Then \[ b+\frac{n}{b} - a-\frac{n}{a} = 1 \implies n = ab - \frac{ab}{b-a}. \]Let $b-a = d$. Since $b-a\mid ab$, we also have $b-a\mid a^2$, so $a^2=de$ for some integer $e$. Now, \begin{align*} n &= a(a+d) - \tfrac{a(a+d)}{d} \\ &= a^2+ad - \tfrac{a^2}{d} -a \\ &= de + d\sqrt{de} - e - \sqrt{de} \\ &= d(d-1) \sqrt{\tfrac{e}{d}} \left( \sqrt{ \tfrac{e}{d}} + 1\right). \end{align*}We claim $\sqrt{\tfrac{e}{d}} \in \mathbb{Z}$. Indeed, we know $\tfrac{n}{b} \in \mathbb{Z}$ since it is a root, so $a-\tfrac{a}{b-a} \in \mathbb{Z}$, so $b-a\mid a$, i.e. $d\mid a$. So $d \mid \tfrac{a^2}{d} = e$. We know $\tfrac{e}{d}=\tfrac{a^2}{d^2}$ is the square of a rational, but since it is also an integer, it must be the square of an integer, as claimed. Therefore, $n$ is of the form \[ n=x(x+1)y(y+1)\]for some $x,y\in \mathbb{Z}$; in particular $x=d-1$ and $y=\sqrt{\tfrac{e}{d}}$. $\blacksquare$ Part (a) We have \[ \sum_{n\in S} \frac{1}{n} = \sum_{x\ge 1}\sum_{y\ge 1} \frac{1}{x(x+1)y(y+1)} = \left( \sum_{x\ge 1}\frac{1}{x(x+1)} \right)^2 = 1, \]which is finite. Part (b) We want to find infinitely many $n$ such that $ab=cd=ef=n$ and $(a+b,c+d,e+f)=(m-1,m,m+1)$ for some $a,b,c,d,e,f,m$. Let \[ n=x(x+1)y(y+1) \quad \text{and} \quad m=2xy+x+y,\]and notice \begin{align*} xy + (x+1)(y+1) &= m+1, \qquad xy \cdot (x+1)(y+1) = n, \\ x(y+1) + (x+1)y &= m, \qquad \qquad x(y+1) \cdot (x+1)y = n. \end{align*}We want to find infinitely many $(x,y)$ such that there exists a $z$ satisfying \[ ( x(y+1)+z ) ((x+1)y -z-1) = n = x(x+1)y(y+1), \]since the former two expressions sum to $m-1$. Expanding the above, it is equivalent to \begin{align*} -xy - xz - x + yz - z^2 - z=0 &\iff (x-z)(y+ (z+1) ) = -2z(z+1). \end{align*}Now, if $x=z-2$ and $y=z^2-1$, then the above equation is satisfied! Backtracking, all $n$ of the form \[ \boxed{n = (z-2)(z-1)(z^2-1)z^2} \quad \forall z\ge 1 \]work for all $z\ge 1$, and the full construction is evident from the above and can be written explicitly.

a) Suppose $n\in S.$ Then, there exist positive integers $a,b,k$ such that $$ab=(a-k)(b+k+1)=n.$$Rearranging yields $$(k+1)a=k(b+k+1).$$Since $\gcd(k,k+1)=1,$ this implies that $$a=k(x+1),\hspace{17pt}b=(k+1)x$$for some positive integer $x.$ Hence, $$\frac{1}{n}=\frac{1}{ab}=\frac{1}{k(k+1)x(x+1)}.$$Therefore, $$\sum_{n\in S}\frac{1}{n}\le\sum_{k,x\ge 1}\frac{1}{k(k+1)x(x+1)}=\left(\sum_{k\ge 1}\frac{1}{k(k+1)}\right)^2=1^2=1,$$as desired. b) Note that $$k(k-1)^2+(k+1)k(k-2)=2k^3-3k^2-k,$$$$k^2(k-2)+(k+1)(k-1)^2=2k^3-3k^2-k+1,$$$$(k+1)(k-1)(k-2)+k^2(k-1)=2k^3-3k^2-k+2.$$Therefore, all $n$ of the form $$n=(k+1)k^2(k-1)^2(k-2)$$satisfy the condition.

For part 1, sum will telescope and you get $$\left(\frac{1}{2}+\frac{1}{6}+\dots\right)\left(\frac{1}{2}+\frac{1}{6}+\dots\right)=1.$$ For part 2, take $n=36x^4+78x^3+54x^2+12x=6x(x+1)(2x+1)(3x+2)$ for $x\geq 1$ with the pairs $(6x^2+7x+2,6x^2+6x),(6x^2+4x,6x^2+9x+3),(6x^2+3x,6x^2+10x+4)$.

To all those claiming that the sum in (a) is exactly $1$ (like the solutions above in #12,#16,#19,#21,#23): It is NOT. This is simply because the double sum $\sum_{a,b \ge 1} \frac{1}{a(a+1)b(b+1)}$ counts most values of $n$ twice (by interchanging $a$ and $b$).

Pensin (A) If $n\in S$, we have that there exists some $m$ so that: $$m^2-4n=t_1^2$$$$(m+1)^2-4n=t_2^2$$ Thus we differentiate into $2$ cases: First case: $m \equiv 1 \pmod{2}$ Then we have that $t_1=2a+1$ and that $t_2=2b$ and that $m+1=2g$, plugging that into our system of equations we have that: $$g^2-n=b^2$$$$g^2-g-n=a^2+a$$from here we have that $b^2-g=a^2+a$, which implies that $g=b^2-a^2-a$, giving us that: $$n=g^2-b^2=(g-b)(g+b)=(b^2-a^2)(b^2-(a+1)^2)=(b-a-1)(b-a)(b+a)(b+a+1)$$if we set $x=b-a-1$ and $y=b+a$, we have that $n=xy(x+1)(y+1)$ thus in this case we have that: $$\sum_{n \in S} \frac{1}{n} \leq \sum_{x=1}^{\infty} \frac{1}{x(x+1)} = \sum_{x=1}^{\infty} \frac{1}{x}-\frac{1}{x+1} < 1$$ Second case: $m \equiv 0 \pmod{2}$ Similarly as in the first case we get that $n=xy(x+1)(y+1)$, leading to the same conclusion. (B) Construction follows from (A).

Claim: $\mathcal F(n) = \{ d + \frac{n}{d} : d \mid n \}$ Proof: Note $m \in \mathcal F(n) \iff m^2 - 4n$ is a perfect square. Write $m^2 - 4n = q^2$ with $q \in \mathbb Z_{\ge 0}$. Then, $$(m+q)(m-q) = 4n$$Note $m = \frac{(m+q) + (m-q)}{2}$. This forces $m+q,m-q$ to have same parity, and since there product is even, so they are both odd. Let $m+q = 2d$, so $m-q = \frac{2n}{d}$. Then $m = d + \frac{n}{d}$. Clerly $m \in \mathbb Z \iff d \mid n$. $\square$ Now assume that $$ \left\vert \left( d_1 + \frac{n}{d_1} \right) - \left( d_2 + \frac{n}{d_2} \right) \right\vert = 1 $$WLOG $d_1d_2 \le n$, otherwise change $d_2 \to \frac{n}{d_2}$. WLOG, $d_1 < d_2$. Now, \begin{align*} 1 &= \left( d_1 + \frac{n}{d_1} \right) - \left( d_2 + \frac{n}{d_2} \right) = (d_2 - d_1) \left( \frac{n}{d_1d_2} - 1 \right) \end{align*}Let $d_1 - d_2 = k$ and $d_1 = y$. Then $d_2 = y+k$ and $$ n = \frac{d_1d_2(k+1)}{k} = \frac{y(y+k)(k+1)}{k} $$Note $y,y+k \mid n \iff k \mid y$. Write $y = kx$. That would force $$n = x(x+1) \cdot k(k+1)$$Conversely, if $n = x(x+1) \cdot k(k+1)$ then divisors $kx,kx+k$ yield two consecutive numbers of $\mathcal F(n)$. We conclude that $$ S = \{x(x+1) \cdot k(k+1) : x,k \in \mathbb Z_{>0} \} $$This implies their infinitude (say, by fixing $x$ and varying $k$). Now, $$ \sum_{n \in S} \frac{1}{n} \textcolor{blue}{\le} \left( \sum_{x \ge 1} \frac{1}{x(x+1)} \right) \left( \sum_{k \ge 1} \frac{1}{k(k+1)} \right)= 1 \cdot 1 = 1$$Note that $\textcolor{blue}{\le}$ cannot be (at least directly) be strengthened to $=$ because of double counting. This solves (a), now we solve (b). We put $d_3 = d_2 + (k+1) = kx + k + (k+1)$ so that we have $$ \left(d_2 + \frac{n}{d_2} \right) - \left( d_3 + \frac{n}{d_3} \right) = (d_3 - d_2) \left( \frac{n}{d_2d_3} - 1 \right) $$This would yield $$ n = \left( \frac{xk + k}{k+1} \right) \left( \frac{xk + k}{k+1} + 1 \right) \cdot (k+1)(k+2) $$We will show for $x = (k+2)(2k+1)$, this value of $n$ and the previous value $x(x+1) \cdot k(k+1)$ are equal. Indeed, \begin{align*} & x(x+1) \cdot k(k+1) = \left( \frac{xk + k}{k+1} \right) \left( \frac{xk + k}{k+1} + 1 \right) \cdot (k+1)(k+2) \\ &\iff x(k+1) = \left( \frac{xk + 2k + 1}{1} \right) (k+2) \\ &\iff x(k+1)^2 = k(k+2)x + (2k+1)(k+2) \\ &\iff x = (k+2)(2k+1) \end{align*}This gives infinite solutions for (b), completing the proof. $\blacksquare$

Claim: If $F(n)$ has two consecutive integers, then $n=ck(c-1)(k+1)$ for integers $c\geq2$ and $k\geq1.$ In order for there to be two consecutive, $n$ must have a factor pair $(a,b)$ and another factor pair $(a+c,b-c+1)$ for some positive integer $c$. Then, $$n=ab=(a+c)(b-c+1)$$$$a(-c+1)+bc+c(-c+1)=0.$$Note that $(0,c-1)$ is a solution. To get another solution from a solution, we need to increase $a$ by $c$ and increase $b$ by $c-1$. Since $c$ and $c-1$ are relatively prime, $$(a,b)=(kc,(k+1)(c-1))$$for an integer $k$ are all the integer solutions to this. Since $a$ and $b$ are positive integers, $k\geq1$ and $c\geq2$. Thus, $$n=ab=kc(k+1)(c-1).$$ This also shows there are infinitely many such $n$, e.g. take $n=2k(k+1)$ for any $k$, which has factor pairs $(2k,k+1)$ and $(2k+2,k)$ whose sum differ by 1. Thus, we clearly have $$\sum_{n\in S}\frac{1}{n}\leq \sum_{c\geq 2}\sum_{k\geq 1}\frac{1}{ck(c-1)(k+1)}.$$We can actually evaluate the sum $$\sum_{c\geq 2}\sum_{k\geq 1}\frac{1}{ck(c-1)(k+1)}$$as follows: $$\sum_{c\geq 2}\sum_{k\geq 1}\frac{1}{ck(c-1)(k+1)}$$$$=\sum_{c\geq 2}\frac{1}{c(c-1)}\sum_{k\geq 1}\frac{1}{k(k+1)}$$$$=\sum_{c\geq 2}\frac{1}{c(c-1)}(1)=1,$$which finishes part (a). For part (b), take $n=36k^4+78k^3+54k^2+12k$.

Epic plane flight headsolve fail Note that by algebraic integers, $p(x)$ must have two rational roots given that it has $1$. Claim: $S$ consists of integers of the form $a(a+1)b(b+1)$. Proof. Two consecutive integers is thus equivalent to $d_2 + \frac{n}{d_2} + 1 = d_1 + \frac{n}{d_1}$ or $(d_2 - d_1) + 1 = \frac{n}{d_1} - \frac{n}{d_2}$. WLOG assume that $d_1 < d_2 \le \sqrt{n}$, from which it follows that \[ \frac{n}{d_1} - \frac{n}{d_2} = n \cdot \frac{d_2 - d_1}{d_1d_2} \ge d_2 - d_1. \]Thus, it follows that $\frac{n}{d_1d_2} = \frac{d_2 - d_1 + 1}{d_2 - d_1}$. As such, it follows that $d_2 - d_1 \mid d_1d_2$. Letting $w = d_2 - d_1$, this becomes $w \mid d_1^2$ and that $n = \frac{w+1}{w} \cdot d_1(d_1 + w)$ which rewrites as $w(w+1) \cdot \frac{d_1}{w} \cdot \left(\frac{d_1}{w} + 1\right)$. By considering $\nu_p$ of $w$ and $d_1$, it follows that $w \mid d_1$ giving the result. $\blacksquare$ Claim: $\sum_{s \in S} \frac1s \le 1$. Proof. This is less than $\sum_{a \ge 0} \frac{1}{a(a+1)} \sum_{b \ge 0} \frac{1}{b(b+1)}$ which telescopes as $1$. $\blacksquare$ Now, take $a = 2x, b = 3x + 2, c = x, d = 6x + 3$. Then note that $(ab, ab + a + b + 1) = (cd + c, cd + d)$. This gives that $n = 2x(2x+1)(3x+2)(3x+3) = x(x+1)(6x+3)(6x+4)$ has $3$ consecutive integers in $\mathcal F(n)$.

Part a. Let $n$ be an element of $S$ and say $n=ab$ and $n=(a-k)(a+k+1)$. Then, \[ab=(a-k)(a+k+1) \iff a(k+1)-bk=k(k+1)\]Thus $k\mid a $ and $k+1\mid b$ from which we obtain that \[a=k(m+1), b=m(k+1)\]for some integer $n$. Thus $n=km(k+1)(m+1)$. From this, we obtain \[\sum_{n\in S}\frac 1n\leq \sum_{a,b\geq 1}\frac{1}{ab(a+b)(b+1)} =\left(\sum_{a\geq 1}\frac{1}{a(a+1)}\right)^2=1\]as desired. Part b. Let $(x,y)$ be such that $2x^2=y^2+1$ where $x,y$ are clearly odd. Then, $2\left(\frac{x^2-1)}{4}\right)=\frac{y^2-1}{4}$. If $x=2a+1$ and $y=2b+1$ then $2a(a+1)=b(b+1)=K$. To finish, note that $K^2=K\cdot K=b^2\cdot (b+1)^2=2a^2\cdot 2(a+1)^2$ which finishes by Pell as \[b^2+(b+1)^2=b(b+1)+b(b+1)+1=2K+1\]and similarly \[2a^2+2(a+1)^2=2a(a+1)+2a(a+1)+2=2K+2.\]Remark: I'm having way too much fun with \paragraph

Part (a) By Vieta's we can write two polynomials, \begin{align*} P_1(x) &= (x - r) \left(x - \frac{n}{r} \right)\\ P_2(x) &= (x - s) \left(x - \frac{n}{s} \right) \end{align*}Note then that we have, \begin{align*} \left(r + \frac{n}{r} \right) + 1 &= \left(s + \frac{n}{s} \right)\\ n\left(\frac{s - r}{rs} \right) &= s - r - 1\\ n &= \frac{rs(1 + r - s)}{(r - s)} \end{align*}Set $r - s = k$ so we have, \begin{align*} n &= \frac{s(s + k)(k + 1)}{k} \end{align*}Note then $k \mid s$ so let $s = mk$ to find, \begin{align*} n &= mk(m + 1)(k + 1) \end{align*}Thus we have, \begin{align*} \sum_{n \in S} \frac{1}{n} \leq \left( \sum \frac{x}{x+1} \right)^2 = 1 \end{align*}thus proven. $\square$ Part (b) Say the three pairs of roots are, $\left( \frac{n}{a}, a \right)$, $\left(\frac{n}{b}, b \right)$, $\left( \frac{n}{c}, c\right)$. Then we require, \begin{align*} a + \frac{n}{a} + 1 &= b + \frac{n}{b}\\ b + \frac{n}{b} + 1 &= c + \frac{n}{c} \end{align*}Dealing with just the first equation we find, \begin{align*} n\left( \frac{1}{a} - \frac{1}{b} \right) &= b - a - 1\\ n\left( \frac{b - a}{ab} \right) &= b - a - 1\\ n &= \frac{ab(a - b + 1)}{a - b} \end{align*}However we can similarly find, \begin{align*} n &= \frac{bc(b - c + 1)}{b - c} \end{align*}Now write $a = b + x$ and $b = c + y$. Then we have, \begin{align*} n &= \frac{b(b + x)(x + 1)}{x} = \frac{c(c + y)(y + 1)}{y} \end{align*}However from the last two equations we have, \begin{align*} \frac{(c + y)(c + x + y)(x + 1)}{x}& = \frac{c(c + y)(y + 1)}{y}\\ \frac{c + x + y}{c} &= \frac{(y + 1)x}{(x + 1)y}\\ \frac{x + y}{c} &= \frac{x - y}{(x+1)y}\\ c &= \frac{(x + 1)(x+y)y}{x - y} \end{align*}So setting $x = y + 1$ we have, \begin{align*} c &= y(y + 2)(2y + 1) \end{align*}works. This gives $b = y(y + 2)(2y + 1) + y$, from which, \begin{align*} n =\boxed{ y (y + 1) (y + 2) (2 y + 1) (2 y^2 + 5 y + 3) } \end{align*}and we're done.

Since $\overline{\mathbb{Z}}\cap\mathbb{Q}=\mathbb{Z}$, $p$ has an integer root iff it has a rational root, which occurs iff $m^2-4n$ is a perfect square. Note that $m^2-4n=t^2\iff (m-t)(m+t)=4n$. Since $m-t$ and $m+t$ have the same parity and $4n$ is even, we must have $m-t=2d$ and $m+t=2\cdot\frac{n}{d}$ for $d\mid n$. Thus $m=d+\frac{n}{d}$ so \[ \mathcal{F}(n)=\left\{d+\frac{n}{d}\ \Big|\ d\mid n\right\}. \]a. For $n\in S$, there exists $d_1,d_2\mid n$ such that $\left(d_2+\frac{n}{d_2}\right)-\left(d_1+\frac{n}{d_1}\right)=1$, which rearranges to \[ (d_2-d_1)(d_1d_2-n)=d_1d_2\tag{1}. \]We claim that $2a(a-1)\in S$ for all integers $a\geq 2$. This is easy: choose $d_1:=a$ and $d_2:=2a$. It follows that $S$ is infinite. Note that (1) rearranges as $n=d_1d_2\left(1-\frac{1}{d_2-d_1}\right)$. Fix $r:=d_2-d_1\geq 2$. Since $\frac{n}{d_2}$ is an integer, it follows that $r\mid d_1$. Thus $d_1=rk$ for a positive integer $k$ so $n=k(k+1)r(r-1)$. Summing over $r$ and $k$, we get \begin{align*} \sum_{n\in S}\frac{1}{n}&=\sum_{r=2}^\infty\sum_{k=1}^\infty\frac{1}{k(k+1)r(r-1)}\\ &=\left(\sum_{r=2}^\infty\frac{1}{r(r-1)}\right)\left(\sum_{k=1}^\infty\frac{1}{k(k+1)}\right)\\ &=1, \end{align*}as desired. $\square$ b. We claim that $a^2(a-1)(a+1)(2a-1)(2a+1)\in S$ for all integers $a\geq 2$. This is easy: choose \begin{align*} d_1&:=(a-1)(a+1)(2a+1)\\ d_2&:=a(a+1)(2a-1)\\ d_3&:=a^2(2a+1). \end{align*}Then $d_1+\frac{n}{d_1},d_2+\frac{n}{d_2},d_3+\frac{n}{d_3}$ are consecutive integers. The conclusion follows. $\square$