?

#### Tannhauser

##### angry insecure male

That is the question, my friend.

1+1=window

#### Tannhauser

##### angry insecure male
Hint for the problem: you can solve it by simply rearranging the equation.

#### Cipher

##### Introspection Specialist
Hint for the problem: you can solve it by simply rearranging the equation.

(Apparently my images got messed up a little, sorry.)

multiply 2 and divide by xy:

substitute y = x + z

multiply (x + z)

$image=http://latex.codecogs.com/gif.latex?2x%20+%202z%20\leq%202x%20+%202z%20+%20\frac{%28y-x%29^{2}}{x}&hash=c5afe1a1a3e73fdb883edbec4eb8a617$

I didn't expect to find a solution by playing around like this. I expected the solution to be much more elegant.

#### Tannhauser

##### angry insecure male
First off, Cipher, you are my hero, as promised. It was a good attempt, but I'd like to point out one thing about your approach: it proves the inequality only for the set in R^2 where xy > 0.

In the inequality
$image=https://latex.codecogs.com/gif.latex?%5Cbg_white%202%20%5Cleq%20%5Cfrac%7Bx%7D%7By%7D%20+%20%5Cfrac%7By%7D%7Bx%7D&hash=827fad39f40a2497aec152569cad8fc2$

you can try to set x=1 and y=-1 and you get 2 <= -2.

So it is only a partial solution . And yes, there is a much more elegant solution as well.

#### Tannhauser

##### angry insecure male
Another hint: start by writing it as
$image=https://latex.codecogs.com/gif.latex?%5Cdpi%7B150%7D%20%5Cbg_white%200%20%5Cleq%20x%5E2%20-2xy%20+y%5E2&hash=fbf2b78782f7529fa22f4a2d9598280f$

#### Cipher

##### Introspection Specialist
Another hint: start by writing it as
$image=https://latex.codecogs.com/gif.latex?%5Cdpi%7B150%7D%20%5Cbg_white%200%20%5Cleq%20x%5E2%20-2xy%20+y%5E2&hash=fbf2b78782f7529fa22f4a2d9598280f$
Yeah, I too just had that epiphany moment during breakfast.

Binomial Theorem:

$image=http://latex.codecogs.com/png.latex?0&space;\leq&space;(x-y)^{2}&hash=49dfb2b54c2695956e20a741944232ed$

Which is true since any real squared is non-negative.

#### Tannhauser

##### angry insecure male
Yeah, I too just had that epiphany moment during breakfast.

Binomial Theorem:

$image=http://latex.codecogs.com/png.latex?0&space;\leq&space;(x-y)^{2}&hash=49dfb2b54c2695956e20a741944232ed$

Which is true since any real squared is non-negative.

Exactly

#### Cipher

##### Introspection Specialist
A guy is lost in the forest. From where he is standing, he can take three different paths. First path takes 2 hours and will lead him home. Second path takes 4 hours and leads back to the same spot. Third path takes 5 hours and also leads back to the same spot.

He picks a path randomly always. What is the expected length of time until he gets home?
Is that one still unsolved?

I'd interpret this as two distinct random experiments.

a) First, our protagonist takes the 4 hour path n times in a row, until he decides to not take it once.

b) After that, he decides between one of the other paths.

a) has a geometric distribution, and since p(break the cycle) = 2/3,
E[number of times the 4-hour-path is taken] = 3/2.
So E1[x] = 4 hours * 3/2 = 6 hours.

In b) E2[x] is obviously the average of 2h and 5h, since p = 0,5.

So in total, E[x] = E1[x] + E2[x] = 6h + 3,5h = 9,5h

#### Tannhauser

##### angry insecure male
@Cipher
It is indeed still unsolved. In your solution, however, it seems that you have assumed that the 3rd path (5 hours) also leads home, but it leads back to the same spot.

#### Cipher

##### Introspection Specialist
Oh, true. I should read more carefully.

In that case, the guy will come to the same spot three (1/p) times on average, choosing both the 4h and the 5h path half of the time on average.

E[x1] = 1,5 * 4h + 1,5 * 5h = 13,5h

After that, the two hours of the remaining path just get added. E[x] = 15,5

#### Tannhauser

##### angry insecure male
I think that reasoning would work for the case with only 1 circular path, but here he might go the 2nd path n times, then 3rd path m times, then the 2nd path some times again and so on.

In any case, 15.5 is not correct

#### Tannhauser

##### angry insecure male
A useful tool for the problem: Law of total expectation

In particular, the special case

$image=https://latex.codecogs.com/png.latex?%5Clarge%20%5Cmathbb%7BE%7D%5BX%5D%20%3D%20%5Csum%5En_%7Bi%3D1%7D%20%5Cmathbb%7BE%7D%5BX%20%5Cmid%20A_i%5D%20%5Cmathbb%7BP%7D%28A_i%29&hash=4e91a227bc4400efa9aafcfad98bdd34$

#### Cipher

##### Introspection Specialist
I think that reasoning would work for the case with only 1 circular path, but here he might go the 2nd path n times, then 3rd path m times, then the 2nd path some times again and so on.

In any case, 15.5 is not correct

Still couldn't figure out why my first intuition was/is or even could be wrong.

Since I think the law of total expectation can be used with my initial approach, I'm gonna stick to it.
The two hours magically get added to the total time after the main experiment is over.
I only look at the expected time it takes the guy to pass all the circular paths he takes until he selects the home path.

Random Variables:
X := number of hours it takes the guy to get home -2
Y := number of times he gets back to the same spot again, excluding the initial situation ( y can be 0)

Events:
Ai := y = i

for large/infinite n:

$image=https://latex.codecogs.com/png.latex?\large%20\mathbb{E}[X]%20\approx\sum^n_{i%3D0}%20\mathbb{E}[X%20\mid+A_i]%20\mathbb{P}(A_i)&hash=cccc61c81bd4798a0d119e81471f12a3$

$image=https://latex.codecogs.com/png.latex?\large%20\mathbb{E}[X]%20\approx\sum^n_{i%3D0}%20\mathbb{E}[X%20\mid+A_i]%20\cdot+\frac{1}{3}%20\frac{2^{i}}{3^{i}}&hash=358a2ec4030e34115a3c3c565bbdad34$

In any event Ai, P[home path is chosen] is 1 at the (i+1). path selection and 0 before that.
=> At any selection before the (i+1). one, P[4h patch] = P[5h path] = 0.5

In the event A0, E[X] = 0.
In the case of A1, E[X] = 0.5 * 4h + 0.5 * 5h = 4.5h

Since any two coin tosses/path selections are independent, every coin toss just adds its own expected result to the total expectancy.
E[Ai] + 4,5h = E[Ai+1]

$image=https://latex.codecogs.com/png.latex?\large%20\mathbb{E}[X]+\approx%20\sum^n_{i%3D0}%20\frac{1}{2}%20\cdot%209%20\cdot%20\frac{2^{i}}{3^{i+1}}&hash=98a775af1592334ca75e86d68cd0eb69$

$image=https://latex.codecogs.com/png.latex?\large%20\mathbb{E}[X]%20\approx\frac{3}{2}%20\sum^n_{i%3D0}%20\frac{2^{i}}{3^{i}}&hash=6a194cd4da67f071aa084beea570526a$

According to Wikipedia, a geometric series like the one described by this sum converges like this

$image=https://latex.codecogs.com/png.latex?\large%20\frac{startvalue}{1%20-%20quotient}%20%3D%20\frac{1}{\frac{1}{3}}%20=%203&hash=236312440e7b72ef4c452f6a63785a84$

=>
$image=https://latex.codecogs.com/png.latex?\large%20\mathbb{E}[X]%20\approx\frac{3}{2}%20\sum^n_{i%3D0}%20\frac{2^{i}}{3^{i}}%20%3D%203%20\cdot+%20\frac{3}{2}%20%3D4.5&hash=6234b942978a723260ad8f77534bbfeb$

Which is... completely impossible. I don't believe this.

Anyways, the solution would be 6.5 hours then.

Edit: Let's try one more time.

A0 := 2h path gets taken at the first try
A1 := 4h path gets taken at the first try
A2 := 5h path gets taken at the first try

$image=https://latex.codecogs.com/png.latex?%5Clarge%20%5Cmathbb%7BE%7D%5BX%5D%20%3D%20%5Csum%5En_%7Bi%3D1%7D%20%5Cmathbb%7BE%7D%5BX%20%5Cmid%20A_i%5D%20%5Cmathbb%7BP%7D%28A_i%29&hash=4e91a227bc4400efa9aafcfad98bdd34$

E[X] = 1/3 * 2h + 1/3 * (4h + E[X]) + 1/3 * (5h + E[X])
E[X] = 1/3 * (11h + 2* E[X])
E[X] = 11h/3 + 2/3 E[X]
1/3 * E[X] = 11h/3
E[X] = 11h

That's... interesting. Now I'm wrong in at least two ways.
I'll give this more thought some other time.

Edit2: I noticed an error in the first solution while dozing off.
I failed to actually make E[X|Ai] dependent on Ai and just put a constant 4.5h. Whoops.
But I'm too tired to correct that now. Might do that later.

#### Tannhauser

##### angry insecure male
Congratulations, Cipher. The last solution you have where E[X] = 11 is absolutely correct. And I think that is the most elegant way of solving it. It's clear and clean.

#### Glaerhaidh

##### straightedgy
Not my invention and some of you might know it. It's a cool one to understand and try to solve.
---
Let a and b be positive integers such that ab + 1 divides a^2 + b^2. Show that

is the square of an integer.

#### Tannhauser

##### angry insecure male
Let a and b be positive integers such that ab + 1 divides a^2 + b^2.

Does that mean that the fraction is an integer?

pi squared

plus 2

squared

#### Glaerhaidh

##### straightedgy
Does that mean that the fraction is an integer?
Yes, look for all the cases when the fraction is such that the division results in an integer.

In the most basic case ab+1 can easily be manipulated to produce 1 and then all products of the following equation would be int.

#### Tannhauser

##### angry insecure male
Yes, look for all the cases when the fraction is such that the division results in an integer.

In the most basic case ab+1 can easily be manipulated to produce 1 and then all products of the following equation would be int.

Interesting problem. But by produce 1, do you mean setting ab+1 = a² + b²? I honestly don't see how that leads to the general case.

#### Glaerhaidh

##### straightedgy
What I want you to do is to prove by listing all cases of when the results of the problem are integers.

Produce 1 meant that denominator would be 1. Denominator =1 isn't the only case which gives correct solutions to the problem. I mentioned it as an example which is also a part of the complete set of possible answers.

I want you to provide me with the complete set of answers or equations/algorithms that describe all of them since listing all the individual numbers for a&b may be impossible and/or tedious.

#### Vrecknidj

##### Prolific Member
(64+4)/(8*2+1) = 68/17 = 4
(900+64)/(30*8+1) = 964/241 = 4
(729+9)/(27*3+1) = 738/82 = 9

Etc.

Excel + Time = Solutions

#### Glaerhaidh

##### straightedgy
Can you provide a general equation for your solutions? This problem is about describing the general case.