[/quote]
,
be real numbers . Prove that : There exist positive integer 
such that 
Where 
be an isosceles triangle 
)
.
passes through 
and 
.
and circumcircle of 
and 
,
be the midpoint of 
and 
be the midpoint of 
.
,
and 
are concurrent.
,
in the array can be changed into either 
or 
so that the row-sums and column-sums remain unchanged. (Note that 
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points 
a non-negative real number 
,
,
.
such that 
and 
are connected by an edge for every 
.
is not fixed and the first selected vertex always has to be 
.![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
,
for which Bobby has a winning strategy.
,
,
,
,
,
be positive integers and let 
.
divides 
and 
.
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
and 
,
and 
meet at 
,
and 
meet at 
Let 
be the point of intersection of the lines 
and 
be the point of intersection of the lines 
and 
Prove that 
is the circle that is tangent to the line segment 
,
be positive reals such that 
.
such that:
and 
: Prove that :
Y by megarnie, TFIRSTMGMEDALIST, Stuart111, Adventure10, deplasmanyollari, Batzorig
Determine all pairs 
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.This post has been edited 2 times. Last edited by djmathman, Dec 29, 2019, 1:43 AM
Reason: edited to reflect actual wording in the shortlist re post #7
Reason: edited to reflect actual wording in the shortlist re post #7
of positive integers for which there exists a positive integer 
and 
or it's immediate (take any 
or 
then no such 
such 
are integers then for any 
there exists 
such that ![\[ \frac{\alpha+\gamma+1}{\beta+\gamma+1} = \frac{M+1}{M}. \]](http://latex.artofproblemsolving.com/c/1/a/c1a36c0bb5200c539eaa250029f095a5f0b0117d.png)
The claim is checked by setting 
.
where 
and 
;
.
for which 
,
.
for 
such that ![\[ \prod_i \frac{\gamma_i + \alpha_i + 1}{\gamma_i + \beta_i + 1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/f/e/5/fe59f107206e0436cb75d90383e8a68a73320413.png)
Similarly we can pick 
for 
such that ![\[ \prod_j \frac{\gamma'_j + \beta'_j + 1}{\gamma'_j + \alpha'_j + 1} = \frac{\ell T+1}{\ell T} \cdot \frac{\ell T+2}{\ell T+1} \cdot \dots \cdot \frac{\ell T+\ell }{\ell T+(\ell -1)} = \frac{\ell T+\ell }{\ell T} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/9/2/4/924d57522f3f953bc60353af7b1624187aab7996.png)
Then if we let 
,![\[ \frac{d(sm)}{d(sn)} = \prod_i \frac{\gamma_i + \alpha_i + 1}{\gamma_i + \beta_i + 12} \prod_j \frac{\gamma'_j + \alpha'_j + 1}{\gamma'_j + \beta'_j + 1} = \frac{T+1}{T} \cdot \frac{T}{T+1} = 1 \]](http://latex.artofproblemsolving.com/c/1/6/c1673220a504409ccfd0769c1641ea0db89136e7.png)
as desired.
and 
instead of 
and 
.
,
,
for any 
so this pair does not work.
,
where 
are integers greater than one such that 
and 
,
are primes. Take a large enough integer 
Choose 
.
Hence
as desired.
,
work. Suppose that 
are the set of primes such that 
,
,
.
such that for all 
,
,
*
.
by our assumption that neither number divides the other.
,
such that 
Call the factor inside the product 
.
,
.
,
such that 
,
,
for some very large 
,
for some very large 
.
.
,
for some very large value of 
,
and 
.
works, so assume 
in what follows. Suppose 
.
.![\[ (x_1+(a_1+1))\cdots (x_k + (a_k+1)) = (x_1+(b_1+1))\cdots (x_k + (b_k+1)). \]](http://latex.artofproblemsolving.com/a/6/d/a6df3569bda08ecff2405917dee26e116c770536.png)
If 
or 
,
or 
.
for 
and 
for 
.
with ![\[ \frac{x_1+b_1+1}{x_1+a_1+1} \cdot \frac{x_2+b_2+1}{x_2+a_2+1} \cdots \frac{x_\ell+b_\ell+1}{x_\ell+a_\ell+1} = \frac{x_{\ell+1}+b_{\ell+1}+1}{x_{\ell+1}+a_{\ell+1}+1} \cdots \frac{x_k+b_k+1}{x_k+a_k+1}. \]](http://latex.artofproblemsolving.com/1/a/d/1adfcc459e4297626199e7abf31adb5eaf433c01.png)
We want to find a solution for the 
'
and similarly,
Now most of the fractions cancel, and we are left with![\[ \frac{x_1+b_1+1}{x_\ell+a_\ell+1} = \frac{x_{\ell+1} + a_{\ell+1} + 1}{x_k+b_k+1}. \]](http://latex.artofproblemsolving.com/2/5/2/252b3bfbb94eedaff40c21477b8b393567f8f541.png)
This clearly has a solution for 
;
,
.
and 
.
works, as choosing any integer 
.
and 
where 
is the set of primes that divide at least one of 
or 
.
,
and 
cannot both be 0. Now, let 
where some 
.
being true. Note that if 
or 
,
always exists to make the equation true (for fixed 
for 
and 
for 
.
Now, to simplify things, suppose we equate 
for all 
,
for all such defined 
At this point, we can just let:
for sufficiently large positive integers 
and 
so that everything is satisfied, and that all 
large enough), and can be easily verified to work (plug back in), so we are done. 
such that neither divides the other. It's easy to see that if 
or 
,
for primes 
and nonnegative integers 
.
for each 
, or 
. Now let us note the following:
with 
and 
such that 
;
(which is positive because 
(for example, 
)
for some large 
or 
,
the canonical factorization of the number 
is the exponent of 
,
is the exponent of 
,
,
and 
to something to get that:
But we must have 
.
we will have that:
and where 
.
was undefined we can practically set 
,
,
Because of the the lemma we have that:
Similarily we have that:
after that point all to be equal to 
,
situation.
and 
.
such that 
For positive integers 
where 
.
where 
and 
.
and dividing, we obtain:
:
as an operation in which a fraction 
is replaced by the product![\begin{align*}
\frac{a}{a+1} &= \frac{a+1}{a+2}\cdot \frac{a^2+2a}{a^2+2a+1}\\[2mm]
&= \frac{b}{b+1}\cdot \frac{c}{c+1}
\end{align*}](http://latex.artofproblemsolving.com/f/b/e/fbe6814ce40d9969dde750110ba43fd98f4f1584.png)
for positive integers 
where 
and 
.
.
:
as an operation in which a fraction 
for positive integers 
where 
.
we can apply the 
or 
respectively, so one will always have more divisors than the other for any 
represent 
where 
.
and 
(all 
and 
are nonnegative integers). We wish to show that there exists nonnegative integers 
,
,
so then we could let 
.
with 
,
such that there exists a positive integer 
.
.
be the set of 
,
.
are irrelevant, so we need only consider the 
This is only possible since 
by replacing every 
with 
,
,
Obviously if 
and 
where 
are primes and 
.
for 
be so that 
.
for 
similarly, and let 
.
Consider the sets 
and 
.
we can just exclude that prime as it doesnt affects the no of divisors of both equally.I claim that 
![\[ \prod_i \frac{x_i + \alpha_i+1}{x_i + \beta_i+1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/c/7/f/c7f57f57b25504c1fa452ebb13f5656e26a681a4.png)
and similarly for set ![\[ \prod_i \frac{x_i + \beta_i+1}{x_i + \alpha_i+1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/1/1/c/11c6829959cf88d439b528cc4d717f9d44ae5bad.png)
Now this 
work, we begin by claiming that for ![\[ \frac{x+a}{y+a} = \frac{N}{N+1},\]](http://latex.artofproblemsolving.com/8/3/5/835adfe02f05bebf1f25c6899d02d2e1422051b5.png)
for all integers 
with 
.
.
,
be sets such that for each prime 
,
,
.
,![\[ \frac{\nu_p (n) + \nu_p (s) + 1}{\nu_p (k) + \nu_p (s) + 1} \]](http://latex.artofproblemsolving.com/8/e/b/8ebcf9d98723abbfec1e74fdf9c9e011f16d89e0.png)
to the ratio the number of divisors of 
so that for each 
for some 
for some 
,![\[ \left(\frac{M|P|-|P|}{M|P|-|P|+1} \cdot \frac{M|P|-|P|+1}{M|P|-|P|+2} \cdot \hdots \cdot \frac{M|P|-1}{M|P|} \right) \left(\frac{M|R|-|R|+1}{M|R|-|R|} \cdot \frac{M|R|-|R|+2}{M|R|-|R|+1} \cdot \hdots \cdot \frac{M|R|}{M|R|-1} \right).\]](http://latex.artofproblemsolving.com/0/3/3/0331f41e975456449248990098680e56f2f392ad.png)
But the left product telescopes to 
and the right telescopes to 
,
.
and for any sufficiently large positive integer 
.
.
is positive.
and 
such that 
and any sufficiently large 
such that for any positive integer 
:
.
.
.![\[\frac{x_1+a_1}{x_1+b_1}\cdots\frac{x_{2^k}+a_{2^k}}{x_{2^k}+b_{2^k}} = \frac{2^{kc}A}{2^{kc}A+1}\frac{2^{kc}A+1}{2^{kc}A+2}\cdots \frac{2^{kc}A+(2^k-1)}{2^{kc}A+2^k} = \frac{2^{kc}}{2^{kc}+2^k}=\frac{2^cA}{2^cA+1}.\]](http://latex.artofproblemsolving.com/d/5/0/d503a0d4159b5752c8a5e45ebf14dd5e579abe5a.png)
.![\[\frac{x_1+a_1}{x_1+b_1}\cdots\frac{x_{2^k}+a_{2^k}}{x_{2^k}+b_{2^k}} = \frac{2^{kc}A+2^k}{2^{kc}A+2^k+1}\frac{2^{kc}A+2^k+1}{2^{kc}A+2^k+2}\cdots \frac{2^{kc}A+(2^{k+1}-1)}{2^{kc}A+2^{k+1}}=\frac{2^{kc}A+2^k}{2^{kc}A+2^{k+1}}=\frac{2^cA+1}{2^cA+2}.\]](http://latex.artofproblemsolving.com/9/4/f/94f25972590de79e47cb0c1413a7e4605821a503.png)
and 
such that 
such that :![\[\prod_{i=1}^{k}\frac{x_i+a_i}{x_i+b_i} = \frac{A}{A+1}\]](http://latex.artofproblemsolving.com/8/e/3/8e30f4aa1b71b0f026b3d99c97a9c251c4f28361.png)
.
where 
are distinct, nonnegative and 
.
![\[\prod_{i=1}^{k}\frac{x_i+a_i}{x_i+b_i} = \frac{2^{c-1}A}{2^{c-1}A+1}\frac{2^{c-1}A+1}{2^{c-1}A+2}\frac{2^{c-2}A+1}{2^{c-2}A+2}\cdots\frac{2A+1}{2A+2} = \frac{A}{A+1}\]](http://latex.artofproblemsolving.com/4/b/8/4b8413fdd349b73e028453e40a72247e0d3873b6.png)
.![\[n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r} p_{r+1}^{a_{r+1}}\cdots p_m^{a_m}\]](http://latex.artofproblemsolving.com/e/2/a/e2a780069630e08dcfcef8b6b98e084ed2dc83dd.png)
![\[k = p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r} p_{r+1}^{b_{r+1}}\cdots p_m^{b_m}\]](http://latex.artofproblemsolving.com/3/8/5/38565f0179e74b053fea181eb5d1cf20b6269fee.png)
![\[s = p_1^{x_1 - 1}p_2^{x_2 - 1} \cdots p_m^{x_m - 1}\]](http://latex.artofproblemsolving.com/d/6/7/d67f876d77d43e5b34e2063f4b46684b2adf3c85.png)
where 
and 
.
.
,
.
if 
,
if 
.
,
so they will clearly have the same number of divisors. Otherwise for 
,![\[X = \sum_{p\in A} v_p(n)-v_p(k) \]](http://latex.artofproblemsolving.com/2/5/f/25f116adc643d3706662cd948c7e58607b87b35f.png)
![\[Y = \sum_{q\in B} v_q(k)-v_q(n) \]](http://latex.artofproblemsolving.com/a/f/f/aff0e5b2de76e5882ff89ebd69b411e521a2c385.png)
for all 
.![\[\prod_{p\in A} \frac{v_p(ns)+1}{v_p(ks)+1} = \prod_{i=1}^{|A|} \frac{MX+\sum_{j=0}^{i} e_i}{MX+\sum_{j=0}^{i-1} e_i}= \frac{MX+X}{MX} = \frac{M+1}{M}\]](http://latex.artofproblemsolving.com/9/9/c/99c6fbacc09e3994e4747f303057a012519bafb3.png)
![\[\prod_{q\in B} \frac{v_q(ks)+1}{v_q(ns)+1}=. \frac{MY+Y}{MY} = \frac{M+1}{M}\]](http://latex.artofproblemsolving.com/0/a/2/0a2ab2b707ed828a06e56fb00db9a945532284e1.png)
![\[\frac{d(ns)}{d(ks)} = \text{(ratio of A-primes)} \cdot \text{(ratio of B-primes)}\cdot \text{(ratio of other primes)} = \frac{M+1}{M}\cdot \frac{M}{M+1}\cdot 1 = 1\]](http://latex.artofproblemsolving.com/2/f/8/2f886d952b366d4f02b5f5ad3c1eea5b81527c6d.png)
and we are done 
for any positive integer 
such that neither one divides the other.
and 
where 
for all 
and 
for all 
.
Our goal is to choose exponents so that the number of factors of 
and 
are the same; i.e. 
Rearrange to get 
inductively for all 
so that 
(To prevent any of the 
arbitrarily large, and we will do that later.) These choices of 
through 
force the numerator of the 
t
Due to the inductive definition of 
for some fixed constant 
and 
,
Notice that 
is a positive integer. Because 
,
.
with appropriate substitutions for 
and fixed positive integer constants 
.
so that 
However, this can be easily done by letting 
Then we can make 
is motivated by the "DURR WE WANT STUFF TO CANCEL" thing, if you don't know what this is then check out vEnhance's functional equation handout 
,
and 
is 
.
where the exponent set of 
,
,
.
,
,
for all 
.
for some positive integer 
.
,
can take on any fraction of the form 
such that 
.
then this becomes 
.
where 
.
for some number 
.
where 
for all 
be top-heavy if 
,
.
are top-heavy 
such that 
and 
where 
and 
into a single top-heavy fraction, and similarly for any two bottom-heavy fractions. The proof follows.
for some positive integer 
such that 
Then 
,
for some positive integer 
and 
,
,
,
,
,
,
or 
when any number is multiplied the greater of ![\[m = p_1^{a_1}p_2^{a_2} \dots p_e^{a_e} \]](http://latex.artofproblemsolving.com/5/5/a/55a8db7bc45e52322271ffdd70aa1418202b7558.png)
![\[n = p_1^{b_1}p_2^{b_2} \dots p_e^{b_e}\]](http://latex.artofproblemsolving.com/a/f/c/afcdb8094c17dae348b26423abea442238aa09bc.png)
![\[s = p_1^{c_1}p_2^{c_2} \dots p_e^{c_e}\]](http://latex.artofproblemsolving.com/1/7/e/17ed110790a175c86c44ecfec073e7f73d0fc893.png)
From the number of divisors of 
we have ![\[ (a_1+c_1+1)(a_2+c_2+1) \dots (a_e+c_e+1) = (b_1+c_1+1)(b_2+c_2+1) \dots (b_e+c_e+1) \]](http://latex.artofproblemsolving.com/9/1/c/91c8a8cf557abf3a3a2e96ad68f7539645de6d59.png)
![\[ \frac{(b_1+c_1+1)(b_2+c_2+1) \dots (b_e+c_e+1)}{(a_1+c_1+1)(a_2+c_2+1) \dots (a_e+c_e+1)} = 1 \]](http://latex.artofproblemsolving.com/a/7/2/a722d0471a6f9fa287a2865c20c135ee50fa9c98.png)
there exist a ![\[ \frac{b+c+1}{a+c+1} = \frac{x+1}{x} \]](http://latex.artofproblemsolving.com/d/f/0/df007103daf75acad524bfb79000c4106c73b440.png)
![\[ \frac{b+c+1}{a+c+1} = \frac{1+x}{x}\]](http://latex.artofproblemsolving.com/e/5/b/e5b4cf0b3f0a289d50155b9ebba57802ccd48e15.png)
![\[ bx+cx+1 = ax+cx+1+a+c+1 \]](http://latex.artofproblemsolving.com/1/2/f/12fc0158f18a06571efab7bdd4f1cd216b660061.png)
![\[ c= x(b-a) -(a+1)\]](http://latex.artofproblemsolving.com/b/3/2/b3203fad5e1fd99409796d39d6ce6e3151c41903.png)
, 
and from 
to 
, 
in such a way such that ![\[ \frac{b_i+c_i+1}{a_i+c_i+1} = \frac{fY+i}{fY+i-1} \]](http://latex.artofproblemsolving.com/c/9/b/c9b945a6495e01da8342facfab0b417f4f33126e.png)
in such a way that ![\[ \frac{a_{f+i}+c_{f+i}+1}{b_{f+i}+c_{f+i}+1} = \frac{(e-f)Y+i}{(e-f)Y+i-1} \]](http://latex.artofproblemsolving.com/9/4/c/94c1c8a7c3495b5e29c971507bc18e977fa7972e.png)
![\[\prod_{i=1}^{e} \frac{b_i+c_i+1}{a_i+c_i+1} = \prod_{i=1}^{f}\frac{b_i+c_i+1}{a_i+c_i+1} \enspace \prod_{i=1}^{e-f} \frac{b_{f+i}+c_{f+i}+1}{a_{f+i}+c_{f+i}+1} \]](http://latex.artofproblemsolving.com/f/5/2/f5219ca5a7b843ef57a4b47470616d21fbc537e9.png)
![\[ = \prod_{i=1}^{f} \frac{fY+i}{fY+i-1} \enspace \prod_{i=1}^{e-f} \frac{(e-f)Y+i-1}{(e-f)Y+i} \]](http://latex.artofproblemsolving.com/f/d/2/fd2b9e64d81e00d3a7a4ff1cc976637011a39015.png)
![\[=1\]](http://latex.artofproblemsolving.com/6/b/b/6bb6af597f731ca95a651de203bf9acb6f51d3a3.png)
.
and 
and 
.![\[ \frac{\alpha_1+\gamma_1+1}{\beta_1+\gamma_1+1} \cdot \frac{\alpha_2+\gamma_2+1}{\beta_2+\gamma_2+1} \cdots \frac{\alpha_m+\gamma_m+1}{\beta_m+\gamma_m+1}=1\]](http://latex.artofproblemsolving.com/f/9/9/f996558662d6a1556aae7e5317bbe27bcace0658.png)
since 
where 
and 
.
for 
and 
for 
(if 
then it is irrelevant since it cancels in our fraction). By 2004 Putnam A1 we see that for every sufficiently large positive integer ![\[ \frac{\alpha_i+\gamma_i+1}{\beta_i+\gamma_i+1} = \frac{N}{N+1}\]](http://latex.artofproblemsolving.com/7/5/c/75c5c34f6e94b1cfb440b8c8dd58909e58b0468a.png)
has a solution for 
and the other way around for 
.![\[ \frac{M}{M+(m-\ell)} \cdot \frac{M+(m-\ell)}{M+2(m-\ell)} \cdots \frac{M+(\ell-1)(m-\ell)}{M+\ell(m-\ell)} \cdot \frac{M+\ell(m-\ell)}{M+\ell(m-\ell-1)} \cdots \frac{M+\ell}{M} =1\]](http://latex.artofproblemsolving.com/d/5/0/d508424fbeff06a95656b3ce2e8f05741f2f9152.png)
where 
and 
.
is a solution, so let 
.
or 
because then 
or 
be the list of all primes, in order from smallest to largest, such that 
Note that if 
will still hold.
and 
then there exists an 
and 
then 
and 
Now, we can add an integer 
and 
so that the products of 
and 
are equal.
,
(and analagously otherwise) solving the equation 
we see there is an integer 
.
:![\[\prod_{i=1}^{j}\frac{1+n_i+r_i}{1+k_i+r_i}=\frac{aN+1}{aN}\cdot \frac{aN+2}{aN+1}\cdot \frac{aN+a}{aN+a-1}\cdot \frac{bN}{bN+1}\cdot \frac{bN+1}{bN+2}\cdot \frac{bN+b-1}{bN+b}\]](http://latex.artofproblemsolving.com/4/c/5/4c5e5bb40d8b3b948767573f4a04ba0c249a4ccd.png)
and we win.
for all 
for all 
denote 
for 
for 
values of 
is irrelevant as it doesn't affect the product of the number of divisors
or 
for all 
.
,
or 
but not 
and at least one 
such that 
(
to both 
,
.
for a large enough 
.
such that the fractions would become 
Therefore, this works if and only if 
and if 
then 
and 
where for at least one 
and for at least one 
Define 
as 
and assume WLOG that for 
we have 
for 
we have 
and for 
we have 
Finally, let 
and 
for 
where 
Then, 
Choose 
for 
Hence, 
as desired. 
If 
,
We want to show that it is possible to choose 
Without loss of generality, say that 
and 
for some 
,
and for 
Then 
such that 
such that 
for any integers 
that are not all the same sign. This is achievable by choosing some constants 
such that the equations 
are true for all 
,
To construct this, without loss of generality let 
be positive. (If it is zero, then take all the 
is negative. Then, set 
,
By choosing 
,
for 
can be expressed in the form 
for some integer 
sufficiently large, they can all be positive integers. Finally, the equation 
will yield a positive integer solution for 
as long as 
Thus, picking 
suffices. As a result, such 
If 
which implies 
Then, notice that 
.
and 
.
where the plus or minus is decided based on which of 
and the other is 
,
Now, we have two possible cases,
Then, 
,
to have 
can be written as the product of some 
rational numbers taking the numerator and denominator as required.
for some prime 
for another prime 
,
![\[ \frac{\alpha+x+1}{\beta+x+1} = \frac{M+1}{M}. \]](http://latex.artofproblemsolving.com/6/1/c/61cad01efdae5ed16f0ce01d0170b59689f770e6.png)
works. 
.![\[ \prod_i \frac{x_i + \alpha_i + 1}{x_i + \beta_i + 1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/b/6/7/b6720b2f3578063ec5d4ee0aba67cc4b92a07038.png)
![\[ \prod_j \frac{y_j + \delta_j + 1}{y_j + \gamma_j + 1} = \frac{\ell T+1}{\ell T} \cdot \frac{\ell T+2}{\ell T+1} \cdot \dots \cdot \frac{\ell T+\ell }{\ell T+(\ell -1)} = \frac{\ell T+\ell }{\ell T} = \frac{T+1}{T}. \]](http://latex.artofproblemsolving.com/c/8/f/c8f14a89bac3e183ed2117527be8b2a187071732.png)
,![\[ \frac{d(sm)}{d(sn)} = \prod_i \frac{x_i + \alpha_i + 1}{x_i + \beta_i + 1} \prod_j \frac{y_j + \gamma_j + 1}{y_j + \delta_j + 1} = \frac{T+1}{T} \cdot \frac{T}{T+1} = 1 \]](http://latex.artofproblemsolving.com/8/a/5/8a52841fa06d0b9434e797dba5d3904721d59b51.png)
and 
,
,
.
to the equation 
Note that 
doesn't matter, since they don't change the product anyway, so let's split all of 
with 
into two groups: 
.
and 
.
Now, this transforms to 
now we aim to make this ``telescope", here's how: First, pick some 
.
,
such that 
.
,
.
,
. So, it suffices for 
,
to this; just pick any one large enough for every 
.
.
be all prime divisors of 
.
We will search 
where we will choose 
Now we have 
Notice that if 
for all 
.
,
so we are done.
for some natural 
and let 
for 
.
Similarly, we obtain 
so we are done.
![\[ m = \prod_i^k p_i^{a_i}\]](http://latex.artofproblemsolving.com/a/e/5/ae5768464bee25d6f5562f4593c5c18fc1d6e363.png)
![\[n = \prod_i^k p_i^{b_i}\]](http://latex.artofproblemsolving.com/b/7/b/b7b19e46da708ead47f712bba381f13a286b9b44.png)
![\[s = \prod_i^k p_i^{c_i}\]](http://latex.artofproblemsolving.com/d/a/4/da4113f3821678f290ff8cfbe80a7ddacbad586b.png)
with 
or vice versa with 
works.
for any 
being defined similarly for 
but 
.
we have![\[\prod_{i=1}^{p} \frac{\alpha_i + c_i + 1}{\beta_i + c_i + 1} = \prod_{i=1}^{p} \frac{p\mathcal{Q} + i - 1}{p\mathcal{Q} + i} = \frac{\mathcal{Q}}{\mathcal{Q} + 1}\]](http://latex.artofproblemsolving.com/8/2/3/823de969e14ff0400da3c72136116c89a371d4f6.png)
and![\[\prod_{i={p+1}}^{k} \frac{\alpha_i + c_i + 1}{\beta_i + c_i + 1} = \prod_{i=p+1}^{k} \frac{(k-p)\mathcal{Q} + i - p}{(k-p)\mathcal{Q} + i - p - 1} = \frac{\mathcal{Q} + 1}{\mathcal{Q}}\]](http://latex.artofproblemsolving.com/0/c/4/0c41a0fec57b60e9abbaf60ff417d31c7e57133c.png)
so we get that 
as desired.
,
,
.
.
is easy. We claim that we can achieve each side of the product being 
for sufficiently large 
for all sufficiently large integers 
,
.
Now when all 
or vice versa without them all being equal it is obviously not possible because every term is 
so the product cannot possibly be 
such that it is in the form 
,
for any integers 
by our claim. Setting 
clearly works so we are done.
if and only if 
and 
,
is a proper divisor of 
.
,
is analogous, so we now show that if 
and 
,
are positive integers, 
are positive integers, and 
are distinct primes. Then, let 
,
are positive integers.
Let 
where 
.
,
and 
cancel in the above equation, and we can ignore the values of 
.
,
,
is a positive integer, and for all 
,
,
is a positive integer. Also, for every valid 
for some positive integer 
.![\[\left(\prod_{i=1}^r \frac{x_i+1}{x_i} \right)\left(\prod_{(c_i,d_i)\in A} \frac{v_i+1}{v_i}\right) = \left( \prod_{i=1}^s \frac{y_i+1}{y_i}\right)\left(\prod_{(c_i,d_i)\in B} \frac{w_i+1}{w_i}\right).\]](http://latex.artofproblemsolving.com/e/6/8/e68d822b12e1e16e6292266a6c58971ee0f39fac.png)
Call a positive rational number that can be written in the form 
for a positive integer 
and 
,
,
,![\[\frac{n+1}{n} = \prod_{i=0}^{m-1} \frac{mn+i+1}{mn+i},\]](http://latex.artofproblemsolving.com/9/7/4/9744b8d57b620626a35c76df0301fd3158fd8ab8.png)
which proves the claim. Now, pick some arbitrary tight number, and express it as a product of 
tight numbers and as a product of 
tight numbers. 
be nonnegative integers for which 
and 
,
for which ![\[\prod_{i=1}^m\frac{a_i+c_i}{b_i+c_i}=1.\]](http://latex.artofproblemsolving.com/d/b/3/db3a216d6a18989d1a2969860a83489756ba375a.png)
,
well always cancel if 
set 
.
and 
,![\[\prod_{i=1}^m\frac{a_i+c_i}{b_i+c_i}=\left(\frac{a_1+c_1}{X+c_1}\right)\left(\frac{a_{u+1}+c_{u+1}}{Y+c_{u+1}}\right).\]](http://latex.artofproblemsolving.com/c/4/6/c465bedac9e0b02d686ec0214beaa9f28cc79e7f.png)
We have 
and 
,
,
(for 
.
)
and 
where 
with 
(if some exponents are equal, cancel them as they dont affect us).
Similarly, 
.
we are done.
,
,
where 
,
,
are prime numbers.
for all 
,
,
such that the following equation holds:
such that 
,
,
where 
.
for some 
for some 
,
to turn these fractions into any value of the form 
and 
,
Since the equation of products is satisfied, we have proven that there exist some choice of 
