# solutions to $\frac{1}a + \frac{1}b + \frac{1}c = \frac{1}{2018}$

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.

• Compare with this post. Aug 20 '20 at 16:45
• math.stackexchange.com/questions/450280/erdős-straus-conjecture/… Aug 20 '20 at 16:52
• COMMENT.-Using the identity $\dfrac 1n=\dfrac {1}{n+1}+\dfrac{1}{n(n+1)}$ we get easily a solution; in fact $$\frac {1}{2018}=\frac {1}{2019}+\frac {1}{2018\cdot2019}=\frac {1}{2020}+\frac {1}{2019\cdot2020}+\frac {1}{2018\cdot2019}$$ Aug 21 '20 at 0:34

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

• Yikes, the prime factors of 2019 is 3 x 673; I am doubtful this question has any "elegant" solutions. Aug 20 '20 at 18:18
• @WillieWong: I get $1296$ for $2019$ and $6884$ for $2020$. $2021=43 \cdot 47$ has no small factors but $2339$ solutions. Aug 20 '20 at 18:22

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 $$\text{gcd}(x,y)>1$$.

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

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

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