Find all Positive Integer Solutions

We find all ways of expressing 1 as a sum of 4 distinct (positive) unit fractions.

00:00 Intro
00:10 Values of a
02:09 Values of b
04:38 Values of c
07:01 Finding solutions

[vvop]

Every Friday night I look forward to another gem by Dr Barker. Thank you.

[MizardXYT]

If you allow duplicates, you also get
1/2 + 1/3 + 1/12 + 1/12 = 1
1/2 + 1/4 + 1/8 + 1/8 = 1
1/2 + 1/5 + 1/5 + 1/10 = 1
1/2 + 1/6 + 1/6 + 1/6 = 1
1/3 + 1/3 + 1/4 + 1/12 = 1
1/3 + 1/3 + 1/6 + 1/6 = 1
1/3 + 1/4 + 1/4 + 1/6 = 1
1/4 + 1/4 + 1/4 + 1/4 = 1

[worldnotworld]

Your capacity to tease puzzles of this sort out rather ordinary mathematics is extraordinary. The fact that it is possible to do so at all reveals just how astonishing mathematics is, and how _playful_ our interaction with it can be. Faced with the original problem here, which seemed very abstract, I had no intuitions. Yet your sober poking about toward a solution, beginning with just plugging numbers in, satisfied every intuition. But more than this: in the end, the very idea that such a puzzle is worthy of public exposition, that there is a class of such worthy puzzles that a mathematician has sought to discover and then to present – it's a bit mind-bending. A meta-mathematics lurks in all this!

[Uranyus]

My immediate solution when looking at the problem is 2, 3, 7, 42 because i know a property of the number 42 which behaves exactly like the problem stated. To my surprise, it turns out there are other (and consecutive too!) solutions to this problem. Great video, doctor. curious if there's some more algebraic solution to this problem rather than primarily guess and check.

[chaosredefined]

There's a useful trick you can use at the end. This is my approach. I did miss out on your trick for eliminating b=5.

1 = 1/a + 1/b + 1/c + 1/d, 0 < a < b < c < d

If a >= 3, then b >= 4, c >= 5 and d >= 6. Therefore, 1/a <= 1/3, 1/b <= 1/4, 1/c <= 1/5, and d <= 1/6. 1/3 + 1/4 + 1/5 + 1/6 = 19/20 < 1, so 1/a + 1/b + 1/c + 1/d < 1. This is a contradiction, so a must be smaller than 3. Therefore, a = 1 or 2. a = 1 means that 1/b + 1/c + 1/d = 0, but this is clearly impossible. Therefore, a = 2, and 1/2 = 1/b + 1/c + 1/d.

If b >= 6, then c >= 7 and d >= 8. Therefore 1/b + 1/c + 1/d <= 1/6 + 1/7 + 1/8 < 1/6 + 1/6 + 1/6 = 1/2. So, 1/b + 1/c + 1/d < 1/2 if b >= 6. This is a contradiction again, so b must be smaller than 6. As it must also be larger than 2, this means that b = 3, 4 or 5.

Suppose b = 3.
1/2 = 1/3 + 1/c + 1/d
Subtract 1/3 from both sides
1/6 = 1/c + 1/d
Multiply through by 6cd
cd = 6c + 6d
Add 36 - 6c - 6d to both sides
cd - 6c - 6d + 36 = 36
(c - 6)(d - 6) = 36
36 is, unfortunately, a highly composite number, meaning it has lots of factors. It can be factored as 1 x 36, 2 x 18, 3 x 12, 4 x 9, and 6 x 6. Since c < d, we don't need to worry about the reverse order of those, and we can discard the last one. From these, we get (c, d) = (7, 42), (8, 24), (9, 18) and (10, 15). We would also get (12, 12), but we can't have them be equal. This gives us: (a, b, c, d) = (2, 3, 7, 42), (2, 3, 8, 24), (2, 3, 9, 18) and (2, 3, 10, 15).

Suppose b = 4
1/2 = 1/4 + 1/c + 1/d
Subtract 1/4 from both sides
1/4 = 1/c + 1/d
Multiply through by 4cd
cd = 4c + 4d
Add 16 - 4c - 4d to both sides
cd - 4c - 4d + 16 = 16
(c - 4)(d - 4) = 16
Playing the same game as last time, 16 can be factored as 1 x 16, 2 x 8 or 4 x 4. That last one is, once again, a dud, so we only get (c, d) = (5, 20) and (6, 12). This gives us (a, b, c, d) = (2, 4, 5, 20) and (2, 4, 6, 12).

Finally, suppose b = 5
1/2 = 1/5 + 1/c + 1/d
Subtract 1/5 from both sides
3/10 = 1/c + 1/d
Multiply through by 30cd
9cd = 30c + 30d
Add 100 - 30c - 30d to each side
9cd - 30c - 30d + 100 = 100
(3c - 10)(3d - 10) = 100
This requires a bit more care than previous ones, so I'm going to go slower. We can factor 100 as 1 x 100, 2 x 50, 4 x 25, 5 x 20, or 10 x 10. Last one can be dismissed coz it would result in c = d. So, we have (3c-10, 3d-10) = (1, 100), (2, 50), (4, 25), or (5, 20). Adding 10 to each, we get (3c, 3d) = (11, 110), (12, 60), (14, 35) or (15, 30). The first and third solutions don't work, as 3c and 3d both need to be divisible by 3. Diving the other two options by 3, we get (c, d) = (4, 20) or (5, 10). But we need c > b, and since b = 5, neither of these solutions work.

So, our solutions are (a, b, c, d) = (2, 3, 7, 42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15), (2, 4, 5, 20) and (2, 4, 6, 12). This should be an exhaustive list.

[ZekeRaiden]

I think my favorite solutions here are the first and last; the first, d = abc, while the last, d=a+b+c.

[BrollyyLSSJ]

One property of the solutions that jumped out to me is that d = lcm(a, b, c).

I can convince myself that it's necessary for all solutions to satisfy this, but as we saw with some cases, it's not a sufficient condition for (a, b, c, d) to be a solution.

[miruten]

This reminds me of looking for uniform tilings of the plane. Then we need 2 = sum_k (1 - 2/n_k) for integers n_k ≥ 3 (or ∞, or even maybe 2 could be allowed). The search pretty simply can be done in the same systemic brute-force approach, and your video gave me a bit of nostalgia for my childhood in-my-head calculations looking at bathroom tiles :)

[GoogleAccount-pict]

Once you get it down to 1/c + 1/d = 1/6 or 1/8, you can use SFFT to solve way faster. (Simons favorite factoring trick)

[yurenchu]

From the thumbnail, I didn't know that integers a, b, c, d had to be distinct and in increasing order, so I found 14 possible combinations (which amount to 205 different solutions):

Assume a≤b≤c≤d .

a = 1 gives no solution (because then RHS>1).
If a > 4, then RHS < 1, so no solutions.
So solutions must be at a = 2, a = 3, a = 4.

a = 4:
1/b + 1/c + 1/d = 3/4 ==> if b > 4, then 1/b + 1/c + 1/d < 1/4, therefore b=4, 1/c + 1/d = 3/4 - 1/4 = 1/2 ==> c = 4, d = 4.
So for a = 4, the solution is {a, b, c, d} = {4, 4, 4, 4}

a = 3 :
1/b + 1/c + 1/d = 2/3 ==> 3 ≤ b < 5
If b ≥ 5, then 1/b + 1/c + 1/d ≤ 1/5 + 1/5 + 1/5 = 3/5 = 0.6 < 2/3, so no solutions.
If b = 4, then 1/c + 1/d = 2/3 - 1/4 = 5/12 ; if c≥5 then 1/c + 1/d ≤ 1/5 + 1/5 = 2/5 = 24/60 < 25/60 = 5/12, so no solution unless c = 4 ==> 1/d = 5/12 - 1/4 = 5/12 - 3/12 = 2/12 = 1/6 ==> d = 6
If b = 3, then 1/c + 1/d = 2/3 - 1/3 = 1/3 ; if c>6 then 1/c + 1/d < 1/6 + 1/6 = 1/3, so no solution unless c = 6, 5, 4 or 3.
==>
if c = 6, then 1/d = 1/3 - 1/6 = 1/6 ==> d = 6
if c = 5, then 1/d = 1/3 - 1/5 = 5/15 - 3/15 = 2/15 ==> d = 15/2 is not integer
if c = 4, then 1/d = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 ==> d = 12
if c = 3, then 1/d = 1/3 - 1/3 = 0 ==> d is not an integer.
So for a = 3, the solutions are {a, b, c, d} = {3, 4, 4, 6}, {3, 3, 6, 6}, {3, 3, 4, 12}

a = 2 :
1/b + 1/c + 1/d = 1/2 ==> 2 < b ≤ 6
If b = 6, then 1/c + 1/d = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 ==> if c > 6, then 1/c + 1/d < 1/3, so c = d = 6
If b = 5, then 1/c + 1/d = 1/2 - 1/5 = 5/10 - 2/10 = 3/10 ==> if c > 20/3, then 1/c + 1/d < 3/10, so c = 5 or c = 6 .
==>
if c = 6, then 1/d = 3/10 - 1/6 = 9/30 - 5/30 = 4/30 ==> d = 30/4 = 15/2 is not integer.
if c = 5, then 1/d = 3/10 - 1/5 = 3/10 - 2/10 = 1/10 ==> d = 10
if b = 4, then 1/c + 1/d = 1/2 - 1/4 = 1/4 ==> if c > 8, then 1/c + 1/d < 1/8 + 1/8 = 1/4, so 4 < c ≤ 8 .
==>
if c = 8, then 1/d = 1/4 - 1/8 = 1/8 ==> d = 8
if c = 7, then 1/d = 1/4 - 1/7 = 7/28 - 4/28 = 3/28 ==> d = 28/3 is not integer.
if c = 6, then 1/d = 1/4 - 1/6 = 3/12 - 2/12 = 1/12 ==> d = 12
if c = 5, then 1/d = 1/4 - 1/5 = 5/20 - 4/20 = 1/20 ==> d = 20
if b = 3, then 1/c + 1/d = 1/2 - 1/3 = 1/6 ==> if c > 12, then 1/c + 1/d < 1/12 + 1/12 = 1/6, so 6 < c ≤12
==>
if c = 12, then 1/d = 1/6 - 1/12 = 1/12 ==> d = 12
if c = 11, then 1/d = 1/6 - 1/11 = 11/66 - 6/66 == 5/66 ==> d = 66/5 is not integer
if c = 10, then 1/d = 1/6 - 1/10 = 10/60 - 6/60 = 4/60 = 1/15 ==> d = 15
if c = 9, then 1/d = 1/6 - 1/9 = 3/18 - 2/18 = 1/18 ==> d = 18
if c = 8, then 1/d = 1/6 - 1/8 = 4/24 - 3/24 = 1/24 ==> d = 24
if c = 7, then 1/d = 1/6 - 1/7 = 7/42 - 6/42 = 1/42 ==> d = 42
So for a = 2, the solutions are {a, b, c, d} = {2, 6, 6, 6}, {2, 5, 5, 10}, {2, 4, 8, 8}, {2, 4, 6, 12}, {2, 4, 5, 20}, {2, 3, 12, 12}, {2, 3, 10, 15}, {2, 3, 9, 18}, {2, 3, 8, 24}, {2, 3, 7, 42}


So there are fourteen possible combinations, and they are:
{a, b, c, d} =
{4, 4, 4, 4}, {3, 4, 4, 6}, {3, 3, 6, 6}, {3, 3, 4, 12}, {2, 6, 6, 6}, {2, 5, 5, 10}, {2, 4, 8, 8}, {2, 4, 6, 12}, {2, 4, 5, 20}, {2, 3, 12, 12}, {2, 3, 10, 15}, {2, 3, 9, 18}, {2, 3, 8, 24}, {2, 3, 7, 42}.

The 205 solutions are found by permuting each of these combinations; they are
(4, 4, 4, 4),
(3, 4, 4, 6), (3, 4, 6, 4), (3, 6, 4, 4), (4, 3, 4, 6), (4, 3, 6, 4), (4, 4, 3, 6), (4, 4, 6, 3), (4, 6, 3, 4), (4, 6, 4, 3), (6, 3, 4, 4), (6, 4, 3, 4), (6, 4, 4, 3),
(3, 3, 6, 6), ... etcetera.

But from the 14 possible combinations, we can easily see that the six only solutions with a<b<c<d are:
(a, b, c, d) = {2, 4, 6, 12}, {2, 4, 5, 20}, {2, 3, 10, 15}, {2, 3, 9, 18}, {2, 3, 8, 24}, {2, 3, 7, 42}.

[holyshit]

We can change the number of variables in this game
For three variables (2, 3, 6) would be a solution
This reminds me another game
Let f(x, y) = (x+y)/(1-xy)
Find a, b in N such that
1=f(1/a, 1/b)
If we want to play with thhee variables equation would be as follows
1=f(f(1/a, 1/b), 1/c)

[AT-zrtv]

Lovely little puzzle, and great explanations.
Have you tried to solve this with 5 such integers, or do the number of cases to evaluate explode? If it's the former, do you see this problem being generalized to N integers, and how would one go about solving it?
Thanks again for all the quality content.

[omograbi]

I remember Michael Penn has solved a similar problem with the use of conjugates, and it seemed so reasonable.

[Foxxey]

I am amazed how many of the left solutions worked. Is there a particular reason why?

[mathcanbeeasy]

The problem is classic. And it does not require the condition of inequality. The condition of inequality is observed and assumed from symmetry, following that any solution (a, b, c, d) must also have all related permutations. This is to make it a little more difficult. If we only want the uniquely ordered solutions, we obviously need the inequality, but this is the biggest clue for the solution.
Then, from the inequality a<=b<=c<=d it follows that 1/a>=1/b>=1/c>=1/d from which it follows that 1/a+1/a+1/a+ 1/a>=1/a+1/b+1/c+1/d=1. From here it follows that 4/a>=1, i.e. a<=4. If you want a, b, c, d to be different, i.e. the inequalities to be strict, then a is even smaller or equal to 3.
And from here we start with the discussion according to the values of a, following that the following equation is identical but only in b, c and d and it is analogous, obviously taking into account the fact that b, c, d must be greater or equal with the previous ones.
Nice problem. Congrats!

[SuperKripke]

2:07 backflips are just a gimmick. Only real mathematicians like Dr Barker shuffle in and out of screen.

[Jkauppa]

optical travelling salesman problem solution, ie, super many photons bouncing all around, if you can adjust the optical path lengths like an old call center

[Deepnil]

Can you do a olympiad series? Love your content

[de_oScar]

I only recently found out about egyptian fractions and read that any positive rational number can be expressed as an egyptian fraction. So far I've only seen how to make egyptian fractions out of proper fractions, but since it's proven that any rational number suffices, *I wonder how to construct natural numbers bigger that 1 in terms of egyptian fractions*.
On one hand we have the fact that the harmonic series diverges, but, on the flipside, no partial sum of the harmonic series is a whole number (other than 1, which is trivial).
You can find egyptian fraction with wolfram alpha, but only for proper fractions (even 13/12 is presented as 1 + 1/12 instead of 1/2 + 1/3 + 1/4).
Of course I could brute force my way through with bounds closer and closer (if 1/2 + 1/3 + 1/4 is too big and 1/2 + 1/3 + 1/5 is too small then… (…)), but it's not mathematically satisfying. Any ideas?

[jursamaj]

Given the amount of case checking involved, a brute force program would be quicker. :)