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
I have not found all possible solutions to this problem
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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:
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: