mmcirvin: (Default)
[personal profile] mmcirvin
"How old is Ann[e]" reminded me of an even older algebra puzzle. This one circulated in 1855 in the letters pages of Robert Merry's Museum, a children's magazine that somehow developed a culture sometimes compared to a Usenet group (or to 20th-century science fiction fandom) in its enormous and chatty letters column. The heated and often silly argument over it paradoxically solidified the letters-column culture as a distinct entity, as these things sometimes do. As Pat Pflieger says:
Like any self-respecting online community, the Cousins indulged in flame wars; though in the Chat flaming often had unexpected consequences: instead of splitting the group, in the Chat it tended to bring readers together. The first flame war, begun in 1855, was central in the development of the column as a virtual community. "That Problem," as it came to be called, looked simple: [solving the system of equations] x2 + y2 = 8; x + xy = 6. To the mathematically inclined, the answer is easy: x = 2; y = 2. To the Cousins, however, the difficulty lay in proving the equation. Proving it to the satisfaction of the other Cousins, that is.
The problem's actually a lot harder than "How old is Ann". As I pointed out in 2002, there's actually a second, non-integer solution in real numbers (and, I think, two more in complex numbers), and the problem is equivalent to solving a quartic equation, which is not easy at all—the second solution can be found analytically but is much easier to find by numerical methods, at least if you're living in the early 21st century.

(I guess I should take the opportunity to say it publicly here, since the spamgates have opened anyway: My current main email address is not the one mentioned there, but matt.mcirvin@gmail.com).

Date: 2008-03-08 01:10 am (UTC)
ext_8707: Taken in front of Carnegie Hall (i think too much)
From: [identity profile] ronebofh.livejournal.com
Hmm, i solved for y and got 7y2 + 16y - 24 = 0. It doesn't have to be quartic, does it? Or is solving for y cheating?

Date: 2008-03-08 01:13 am (UTC)
From: [identity profile] mmcirvin.livejournal.com
y=2 doesn't satisfy that equation, so I think something's wrong.

Date: 2008-03-08 01:17 am (UTC)
ext_8707: Taken in front of Carnegie Hall (oops)
From: [identity profile] ronebofh.livejournal.com
Oops, it should be -28, not -24. Simple arithmetic eludes me.

Date: 2008-03-08 01:19 am (UTC)
From: [identity profile] mmcirvin.livejournal.com
y=2 doesn't solve that equation either.

Date: 2008-03-08 02:06 am (UTC)
ext_8707: Taken in front of Carnegie Hall (grumpy)
From: [identity profile] ronebofh.livejournal.com
I think i'd better quit while i'm behind...

Date: 2008-03-08 05:38 pm (UTC)
From: [identity profile] ikkyu2.livejournal.com
You can't simplify this to a quadratic. You've done something impermissible, probably to do with ignoring one of the real roots of an even exponent.

Date: 2008-03-08 07:08 pm (UTC)
ext_8707: Taken in front of Carnegie Hall (oops)
From: [identity profile] ronebofh.livejournal.com
No, i've simply done bad algebra. I did it in a hurry on a Post-It at work.

Date: 2008-03-10 02:57 am (UTC)
From: [identity profile] eb-oesch.livejournal.com
Dabbler! Dilettante! I took the two equations and successfully solved for z.

Date: 2008-03-10 04:52 am (UTC)
ext_8707: Taken in front of Carnegie Hall (cornholio)
From: [identity profile] ronebofh.livejournal.com
POST PROOF OR RETRACT

Date: 2008-03-08 08:45 am (UTC)
From: [identity profile] smashingstars.livejournal.com
IHNJH; IJLS "Dick Boldhero."

Of course,

Date: 2008-03-08 01:20 pm (UTC)
From: [identity profile] notr.livejournal.com
since you already know one root by guessing, you can divide that one out and you're just left with a cubic equation to solve. Still not a trivial solution, but it had been known for 300 years.

Re: Of course,

Date: 2008-03-08 01:21 pm (UTC)
From: [identity profile] mmcirvin.livejournal.com
That's true.

Re: Of course,

Date: 2008-03-09 01:49 pm (UTC)
From: [identity profile] mmcirvin.livejournal.com
...This in turn raises interesting questions about what constituted "proving an answer" in the minds of the young correspondents, which I guess I'd have to read all the letters to discover. Of course, if you want to check that x=2, y=2 is a correct answer, you can do that with five seconds of mental arithmetic. But it must have seemed somehow illegitimate to a veteran of grammar-school arithmetic drill to get an answer to a problem by guessing and then checking the answer; they'd have wanted a turn-the-crank procedure for extracting solutions from this class of problems. And in a sense that's true, since just guessing and checking won't prove that that is the unique answer--particularly in this case, since it isn't the unique answer!

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