Online Course Discussion Forum
MCIII Number Theory 2.28
Hello,
I found that it must be 3 since the sum that is subtracted from 4 cannot be greater than 1. I also found 1/2+2/3+8/9+17/18=3, but how do I find all of them?
Thanks,
Tina Jin
$n=3$ is correct, and the example you found is good.
To find all the sets, let $a < b < c < d$ and $$\frac{a-1}{a} + \frac{b-1}{b} + \frac{c-1}{c} + \frac{d-1}{d}=3.$$ So it turns out that $$\frac{1}{a}+\frac{1}{b} +\frac{1}{c} + \frac{1}{d}=1.$$
The choices for $a$ are very limited. Similar for $b$. Once you fix $a$ and $b$, you can solve for $c$ and $d$.
I still have a question on this problem:
Here's what I did:
Note a=2 and only 2 because if a≥3, then 1/b+1/c+1/d≥2/3 but max(1/b+1/c+1/d)<2/3.
I fixed a=2:
Fix b=3:
1/c+1/d=1/6
Expand and factor: (c-6)(d-6)=36, then get corresponding solutions for corresponding factor pairs.
And the same for b=4, 6 (b=5 doesn't give integers c and d). But some solutions overlap.
How do I know if a value of b will give an integer, and how do I know if a solution will overlap? Do I have to test everything out until the biggest number in the variable set (42 I think, since 1/2+2/3+6/7+41/42=3 and 41/42 is the closest solution I got to 1)
Thanks,
Tina Jin
Each solution is an ordered quadruple $(a,b,c,d)$, so there is no such thing as "overlap" of solutions.
There is only one case for $a$: $a=2$. There are only two cases for $b$: $b=3$ and $b=4$. Then for each case you solve for $c$ and $d$, and only a few choices in each case. You are already on the right track of finding them, so why not just carry it out?
I am still a little bit confused. How do you know there can only be two cases, b=3,4?
By Simon's Favorite Factoring Trick,
If b=5: 1/c+1/d=3/10
10c+10d=3cd so 3cd-10c-10d=0, so (c-10/3)(d-10/3)=10/9. Now multiply both sides by 9 to get (3c-10)(3d-10)=10 c=4 and d=5 work, but now d=b so there are no solutions.
But of course it is not possible to do this for every integer b>5.
How do you prove it must be b≤4?
Thanks,
Tina Jin
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