I enjoyed that bit. I clever, funny bit of writing. Lisbeth is a great character and her brilliance at math adds dimension.

Posted by Tesserae in North Bay ,

I agree that this may have been overegging it a bit. However because of the amazing plot and awesome heroine I can easily overlook this.

Posted by astrol in Dover,

I still could not figure what was the solution that Salander came to. Did you guys get any hint? It just says that she got the solution, but I am confused and wondering what triggered the solution to her at that moment.

Posted by Kaushal in Hillsborough, NC ,

I’s still lost with the whole theorem…may be you genii may care to explain it to the laymen

Posted by emililab in sydney ,

This is where you nod your head sagely, and if anything, “Oh” is the proper response. As if you ever heard of Fermat or cared if his theorem was solved. The same way you do when someone mentions E=MC2, or doppler.

Posted by sgpatien in livermore, ca ,

im not good in reading english………….but this book is great!!!!!!!!!!!!!!its hard to put it down!!!!!

I like a lot the exaplanation in the 3rd book of how Salander cannot remember the solution. If you recall the bullet penetrated in a part of the brain that is responsible for mathematical capacity (well so it says in the book, I could not know for myself). I dont expect that the “intuitive” solution (not the actually existing on) to that would be provided in any of the 10 books, (except if S.L. solved the Fermat problem on his own in the same manner).

Posted by UR in Heraklion, Crete ,

personally being lousy at maths i came to something of a conclusion…

3^2 4^2=5^2 All nice and kosher. It is however 2 dimensional

it cannot work 3d, or ^3 because its just two cubes. With 3 however, the solution is admirably beautiful:

3^3 4^3 5^3=6^3

So basically adding one more numer, and in the right damn order 1,2,3,4,5,6,7,8…

I do wonder if my lil headcalculated “proof / game” holds for th 4th, 5th and so forth..

Posted by ikks in helsingfors, Finland ,

It ain’t that easy. See http://en.wikipedia.org/wiki/Wiles'_proof_of_Fermat's_Last_Theorem

Posted by thnidu in Philadelphia, PA, USA ,

You don’t need an in depth knowledge of this to understand Salander’s solution, and as she says, a philosopher would have more chance – it is a riddle, a rebus, you “line it up” and it is simple and the answer is funny: a rebus is a play or pun (roughly) and if you line up the z’s

x³ y³=z³

in other words

x³ y³=zzz and zzz=sleep

an unsolvable maths problem puts you to sleep, and the higher the exponent the more the damn puzzle puts you to sleep ie the higher the exponent the more z’s you have.

But…. the funniest part of the joke is that there is also a very subversive humour in the fictional “fact” that all these self-important, balding, grey-haired, crusty, walk-short, knee-high socks ‘n’ sandals, cardigan and pocket-protector wearing (insert stereotype here) mathematical geniuses have spent years and years trying to solve/prove a theorem when all along, Fermat has been (fictionally) “taking the piss”, and that his theorem is in fact a joke or riddle or rebus, but instead of being able to see it, he’s had all these mathematicians busting a gut over something which they are (by their mathematical nature) too blind to see. ( I don’t know if Americans use the term “taking the piss” but it is basically having a subversive or mocking joke at someone else’s expense ).

Salander would have liked that.

That is why Salander giggles and why she calls him a cocky devil

…..and the more Zs you have, the less likely it is to fit in the margin…..

No wonder she had to sit down.

Posted by Jess in NZ ,

Thank you Jess for explain it!

Posted by Luz in Domican Republic ,

i came up with one cubed, minus negative one cubed, equals zero cubed.

Posted by kiteless in atl ,

Brilliant solution, kiteless, but it would be “plus” not minus, since negative one cubed equals negative one: (1)^3 (-1)^3 = (0)^3. This would also work for all odd exponents from 1 to infinity. Unfortunately, it doesn’t fit Fermat’s rules that the x, y and z have to be positive non-zero integers. DANG!

Brilliant solution, nonetheless. Hat tip to you.

Posted by rickbull in Nashville ,

In further considering your solution, kiteless, I find that this solution works for any values of x and y so long as they are positive and negative integers of the same value, and z would always be zero. Thanks for getting my brain working on a Saturday. Feels good.

Posted by rickbull in Nashville ,