Hey all. Is there anyway to find ALL such integers a,b,c,d such that a<b<c<d and
1/a + 1/b + 1/c + 1/d = 1?
I know a = 2 since if a = 3 or more then the sum will be less than one. After that I'm stuck.
Any help is appreciated.
Thanks
Hey all. Is there anyway to find ALL such integers a,b,c,d such that a<b<c<d and
1/a + 1/b + 1/c + 1/d = 1?
I know a = 2 since if a = 3 or more then the sum will be less than one. After that I'm stuck.
Any help is appreciated.
Thanks
Hello, knockouwt4!
$\displaystyle \text{Is there anyway to find ALL such integers }a,b,c,d$
$\displaystyle \text{such that }\,a<b<c<d\,\text{ and }\,\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} \:=\:1\,?$
I found four solution so far:
. . $\displaystyle \displaystyle{ \frac{1}{2} + \frac{1}{3} + \frac{1}{9}+ \frac{1}{18}\;=\;1\qquad\qquad\frac{1}{2}+\frac{1} {4}+\frac{1}{5}+\frac{1}{20} \;=\;1$
. . $\displaystyle \displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24} \;=\;1 \qquad\qquad \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42}\;=\;1 $
It can be done for three integers: .$\displaystyle \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} \;=\;1$
and five integers: .$\displaystyle \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{14} + \frac{1}{28}\;=\;1$
and nine integers: .$\displaystyle \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{31} + \frac{1}{62} + \frac{1}{124} + \frac{1}{248} + \frac{1}{496}\;=\;1$
I'll let someone else explain where my denominators are coming from . . .
I have not found all possible solutions to this problem
-----------EDIT--------------------
If a = 2 and b = 3 are fixed, the way to solve that part is to reduce it to:
6(a+b) - ab = 0
(a-6)(6-b)= 36
Now we can look at various factorizations of 36 and solve for (a,b) by a case by case analysis.
-----------EDIT-------------------------
However the previous posts (including the OP's post) are assuming that the integers are positive. The question allows negative integers too and here is a possible solution:
$\displaystyle \frac{1}{-6}+\frac{1}{-3}+\frac{1}{1}+\frac{1}{2} = 1$
We will have to include these solutions too . .
Okay if I didn't make a mistake then this is the full list, gives b, c, d
PARI/GPCode:3 7 42 3 8 24 3 9 18 3 10 15 4 5 20 4 6 12
Edit: Didn't think of negative integers.Code:f()=for(b=3,5,for(c=b+1,11,d=1/2-1/b-1/c;if(d>0&&numerator(d)==1&&denominator(d)>c,print(b," ",c," ",denominator(d)))))
Trying to move towards a complete solution: if not all of a,b,c,d are positive then either
(a,b,c,d) = (-12,2,3,4)
(a,b,c,d) = (-30,2,3,5)
or b = 1 or c = 1.
There is an infinite class of solutions
(a,b,c,d) = (-6k, -3k, 1, 2k)
where k ranges over the positive integers.
Edit: More solutions
(-2, 1, 3, 6)
(-3, 1, 4, 12) (-12, -4, 1, 3)
(-4, 1, 5, 20) (-20, -5, 1, 4)
(-4, 1, 6, 12)
(-5, 1, 6, 30) (-30, -6, 1, 5)
(-6, 1, 7, 42) (-42, -7, 1, 6)
(-6, 1, 8, 24) (-24, -8, 1, 6)
(-6, 1, 9, 18)
(-6, 1, 10, 15) (-15, -10, 1, 6)
(-7, 1, 8, 56) (-56, -8, 1, 7)
(-8, 1, 9, 72) (-72, -9, 1, 8)
(-8, 1, 10, 40) (-40, -10, 1, 8)
(-8, 1, 12, 24)
So the solution to this problem revolves around the question, find all pairs (u,v) of distinct positive integers such that 1/u + 1/v is a unit fraction; in other words, u+v divides uv.
There's a nice writeup by euler (post # 8) in the solution forum to this problem on Project Euler
http://projecteuler.net/index.php?se...roblems&id=108
but the solution forum is restricted to those who've solved that problem and entered the solution. Of course we would discard solutions of the form 1/8 + 1/8 = 1/4 but everything else would apply.
Spoiler: