Special Teaser

Here is a problem by j.a.h. Hunter.

Our Knight has spelt a neat little teaser for us today; you'll see what it is if you follow his moves on this part of a chessboard.

HIS MORE AND TIMES AGES ITSELF ITSELF TIMES TWELVE HER THAN AGE AGE THEIR HIS ADD TIMES WHAT AGE TO SO HER UP AGE ARE

Please express the teaser explicitly and then solve it.

Peter Heichelheim

spoiler space

A 'knight's tour' of the grid spells out the teaser as follows:

19 14 5 8 25 4 9 18 13 6 15 20 7 24 1 10 3 22 17 12 21 16 11 2 23

'His age times itself and her age times itself add up to twelve more than her age times his age. So what are their ages?'

x^2 + y^2 = 12 + xy x^2 - xy + y^2 - 12 = 0

The only positive solutions for this are x=4, y=2 or x=2, y=4. Courtesy Dario Alpern's quadratic Diophantine equation solver:

Spoiler below quoted text. Peter Heichelheim steals another problem from J.A.H. Hunter:

The relevant knight's tour is e3-d1-b2-a4-c5-e4-c3-d5-b4-a2-c1-e2- d4-b5-a3-b1-d2-c4-a5-b3-a1-c2-e1-d3-e5, or if you prefer:

19 14 5 8 25 4 9 18 13 6 15 20 7 24 1 10 3 22 17 12 21 16 11 2 23

producing 'His age times itself and her age times itself add up to 12 more than her age times his age. So what are their ages?'

That is, we need a solution in positive integers to:

x^2 + y^2 = 12 + xy

If x and y were both odd, we'd have odd + odd = even + odd; and if one was odd and one was even, we'd have odd + even = even + even. Both are impossible, so x and y are both even. Also, we have (x + y)^2 = x^2 + y^2 + 2xy = 12 + 3xy, which is divisible by 3, so x+y must also be divisible by 3.

The smallest possible solution for x and y would therefore be 2 and 4... and this proves to work. By trial and error there is no other solution in small numbers; and once the numbers stop being small, the squared terms on the left dominate, so there is no other solution at all. 'Their ages' are 2 and 4.

A budding mathematician once told her father an interesting fact: 'Did you know that if you subtract the product of our ages from the sum of the squares of our ages, you'll get the current year?'

The father was suitably impressed. Ten years later, he reminded his daughter of the mathematical coincidence. She responded: 'Well, that's obviously not true this year. But come to think of it, if you happened to be the same age you were ten years ago, then the coincidence would hold again this year!'

The father and daughter are still alive today. How old are they now?

Spoiler below quoted text. Ben Zimmer writes:

There is insufficient information to provide an exact answer.

21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

At the time of the daughter's original remark, call her age x, her father's age z, and the year y. Then

x^2 + z^2 - xz = y

And 10 years later, we have

(x+10)^2 + z^2 - (x+10)z = y + 10

That is,

x^2 + 20x + 100 + y^2 - xy - 10y = z + 10

Subtracting the first equation, we get z = 2x + 9, and therefore y = x^2 + (2x + 9)^2 - x(2x + 9) = 3(x^2) + 27x + 81. By experiment we find that

if x = 20 then y = 1821 if x = 21 then y = 1971 if x = 22 then y = 2127

Clearly y increases monotonically as (positive) x increases, so the only possibility that fits the problem statement is y = 1971. Then x = 21 and z = 51.

If the date of the statements made in 1971 and 1981 was known to be the same as today's date in 2002, then the answer would be that the daughter is now 52 and the father 82, but we don't have that information. Either or both of the two could be either a year younger or a year older than those ages.

[snip solution]

[spoiler space]

FS + F^2 + S^2 = Y 3(F+3) + (F+3)^2 + 9 = Y+3

F^2 + 9F + 24 = Y

S^2 + FS - (9F+24) = 0

-F + sqrt(F^2 + 36F + 96) S =

Before Mark B. points out that there is insufficient information to provide an exact answer, let me change the question to:

In what year was the speaker's sister born?

You and Ben Zimmer got J.A.H. Hunter's answer. I am curious did you or Ben Zimmer figure out the knight's tour by looking at the words and figure out the teaser from that or did something more. After all there seem to be quite a few combinations.

Yup.

F^2 +36F + 96 = (F+18)^2 - 228 = P^2 (say)

Therefore (F+18)^2 - P^2 = 228 = 3*4*19

That is, (F+18+P)(F+18-P) = 3*4*19

The two factors on the left have the same parity (they differ by 2P), so they are both even. Assuming F,P>=0, F+18-P = 2 or 6. This gives sqrt(F^2 + 36F + 96) is an integer when F=40 or 4.

[9-S]F=[S^S-24]=[S-9][S+9]+57 F=57/[9-S] - [S+9] 57=19*3*1, so (9-S), 6 or 8 , ( or -10, - 66 ) if S=6, F=4 S=8, F=40.