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Find, with proof, all ordered triplets of positive integers $(a,b,c)$ so that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}{2018}.$

In general, if $d$ is a positive integer, then $(a,b,c) = (3d,3d,3d),(d,2d,6d),(d,6d,2d), (2d,d,6d),(2d,6d,d),(6d,d,2d),$ and $(6d,2d,d)$ are all ordered triplets of positive integers such that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}d.$ However, I'm unsure how to find all such triplets. I tried manipulating the equation as follows: \begin{align} &1 + \dfrac{a}b + \dfrac{a}c = \dfrac{a}{2018}\\ &\dfrac{a}b + \dfrac{a}c = \dfrac{a-2018}{2018}\\ &\dfrac{ab}{c } = \dfrac{b(a-2018)-2018a}{2018}\\ &c[b(a-2018)-2018a]-2018ab = 0\\ &c[(a-2018)(b-2018)-2018^2]-2018ab = 0\\ &c[(a-2018)(b-2018)]-2018(ab-2018(a+b)+2018^2+2018(a+b)-2018^2)-2018^2 c =0 \\ &(c-2018)(a-2018)(b-2018)-2018^2(a+b+c)+2018^3 = 0\\ &(c-2018)(b-2018)(a-2018) = 2018^2(a+b+c) - 2018^3, \end{align}

but I don't know whether this is useful.

Source (from comments): I based this off a contest problem. I came up with it by myself. The question I based it on was the Putnam 2018 question A1.

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I wrote a little Python program that did the brute force. It found $670$ solutions with $a \le b \le c$ The first was $$a=2019, b= 4074343, c= 16600266807306$$ There were $40$ with $a=2019$ and the last of those was $$a=2019, b=c=8148684$$

As long as nobody laughs too loud, here is the code. I just let $a$ range from $n+1$ to $3n$, compute the range of $b$ to be so that $\frac 1b \le \min(\frac 1a, \frac 1n-\frac 1a)$, compute $c$ using integer division, then see if it comes out even. succ counts the successes. This is Python 2.

def prog(n, plev=0):
    succ=0
    astart=n+1
    aend=3*n
    if (plev > 19): print 'astart, aend',astart, aend
    for a in xrange(astart,aend+1):
        bstart=max(n*a/(a-n)+1,a)
        bend=2*n*a/(a-n)
        if (plev > 19):  print 'bstart, bend', bstart, bend
        for b in xrange(bstart, bend+1):
            c=n*a*b/(a*b-n*(a+b))
            if (n==a*b*c/(a*b+a*c+b*c)):
                print 'success',a,b,c
                succ+=1
    print 'successes',succ
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  • $\begingroup$ Yikes, the prime factors of 2019 is 3 x 673; I am doubtful this question has any "elegant" solutions. $\endgroup$ Aug 20 '20 at 18:18
  • $\begingroup$ @WillieWong: I get $1296$ for $2019$ and $6884$ for $2020$. $2021=43 \cdot 47$ has no small factors but $2339$ solutions. $\endgroup$ Aug 20 '20 at 18:22
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Partial answer. $\,$ Let $d=\text{gcd}(a,b,c)$ and, thus, $a=dx, b=dy, c=dz$ - where $\text{gcd}(x,y,z)=1$. If we assume that $\text{gcd}(x,y)=\text{gcd}(y,z)=\text{gcd}(z,x)=1$ then $\text{gcd}(xyz, xy+yz+zx)=1$ $$\frac1a+\frac1b+\frac1c=\frac1{2018}\iff \frac{abc}{ab+bc+ca}=\frac{dxyz}{xy+yz+zx}=2018$$ This implies that $xy+yz+zx\mid d\iff d=(xy+yz+zx)\cdot k, k\in\mathbb Z$. Hence, the equation becomes $$k\cdot xyz=2018$$ Once you have the integer solutions to this equation - and this should not take long, since $2018=2\cdot 1009$ -, use the values to obtain back the solutions $$(a,b,c)\equiv (kx\cdot (xy+yz+zx),ky\cdot (xy+yz+zx), kz\cdot (xy+yz+zx) )$$

Observation. Due to symmetry, you "only" have to consider the case $\text{gcd}(x,y)>1$.

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$\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}{2018}\implies \Big(a b - 2018 (a + b)\Big) \Big(a c - 2018 (a + c)\Big)=\Big(2018a\Big)^2$

Let $a<b<c$ and $\Big(2018a\Big)^2=mn$, where natural $m<n$.

Then $b=\dfrac{2018 a + m}{a-2018}$ and $c=\dfrac{2018 a + n}{a-2018}$

pari/gp code:

abc2018()=
{
 s= 0;
 for(a=2019, 10^4,
  d= (2018*a)^2;
  D= divisors(d);
  for(i=1, #D\2,
   m= D[i];
   b= (2018*a+m)/(a-2018);
   if(a<b, if(b==floor(b),
    n= d/m;
    c= (2018*a+n)/(a-2018);
    if(b<c, if(c==floor(c),
     if(1/a + 1/b + 1/c == 1/2018,
      s++;
      print("("a", "b", "c")")
     )
    ))
   ))
  )
 );
 print("\nNumber of solutions: "s)
};

Output:

(2019, 4074343, 16600266807306)
(2019, 4074344, 8300135440824)
(2019, 4074345, 5533424985330)
(2019, 4074346, 4150069757583)
(2019, 4074348, 2766714529836)
(2019, 4074351, 1844477711338)
(2019, 4074354, 1383359302089)
(2019, 4074360, 922240892840)
(2019, 4074378, 461122483591)
(2019, 4075015, 24670140810)
(2019, 4075351, 16456267338)
(2019, 4075688, 12337107576)
(2019, 4076360, 8230170840)
(2019, 4076361, 8226096498)
(2019, 4077034, 6170590959)
(2019, 4077369, 5488138674)
. . .
(3636, 4545, 2038180)
(3687, 4458, 921365323)
(3700, 4440, 22399800)
(3828, 4268, 187328922)
(3885, 4200, 22399800)
(3940, 4137, 83484660)
(3960, 4115, 657682344)
(4002, 4071, 39707177)
(4036, 4037, 16293332)
(4036, 4038, 8148684)
(4036, 4040, 4076360)
(4036, 4044, 2040198)
(4036, 4052, 1022117)
(4036, 5045, 20180)
(4036, 6054, 12108)

Number of solutions: 658
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