Herer is his solution again.

The first result is moderately interesting:

(a+b)^2 + (a-b)^2 = 2 (a^2 + b^2)

But the second one is then trivial

(a+b)^2 + (a-b)^2 + a^2 + b^2 = 3 (a^2 + b^2)

Do you have some extra constraints for the second problem to make it non-trivial?

For most folks of my generation, raised on lots and lots of factoring problems, the first problem was one of a pair that jumped off the pages of algebra books so often they were almost memorized. The sum and difference of (x+y)

^{2}and (x-y)

^{2}produce the results 2( x

^{2}+y

^{2}) and 4xy; both of which seemed to occur in textbook problems of my youth.

Had Gas... not already told me he had always had problems with memory I would think he had grown up on the same factoring fare and had also memorized them. Perhaps he is more like Dodgeson (Carroll) who apparently wrote about re-discovering this in his diary "towards the end of his life" as he too did not recall it. As a puzzle maker, he probably thought the addition of a third multiple would disguise a common factoring and it was this second problem that appeared in his "Pillow Problems".

Since my own memory also is very fragile, I was fortunate to find the information in Edward Wakeling's "Lewis Carroll's Games and Puzzles".

Carroll also included a somewhat more difficult probability problem in his "Pillow Problems". Since I can't make the first problem more difficult for Gas.... I will give the advanced version of this one.

An urn holds B Black and W White balls. One is selected at Random and placed in a bag without your knowledge. Then a White ball is added to this bag. A random draw from this bag produces a white ball. What is the probabilty the original ball was White?

## 4 comments:

Me, me, me... are we supposed to post spoilers here?

Yep, Post away....

I posted a spoiler yesterday, but it seems not to have made it to the blog for some reason. I'll try again.

This is Bayes' Rule.

We have two possibilities: WW and BW.

We have two possible observations: x=White and x=Black.

We want P(WW | x=White), which is

P(x=White| WW) P(WW) / P(x=White)

The prior probability P(WW) = W/(B+W)

The likelihood P(x=White|WW)=1

The probability of the observed event P(x=White)= P(WW)P(x=White|WW) + P(BW)P(x=White|BW)

= (W+0.5B)/(B+W)

So our posterior probability P(WW|x=White) = W/(W+0.5B)

On the previous problem, I'm ashamed to admit that I didn't immediately recognize the result and thought it was supposed to be a geometry problem. After wasting 2 or 3 minutes drawing triangles, I looked at your example using 7 and 4 to get 11 and 3. The sum and difference then popped out at me, and I did a quick check to see that the produced the desired result. I should have known to look at the example sooner.

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