Problem 7, April 1996Author(s): Robert H. BeckerSource: The Mathematics Teacher, Vol. 89, No. 6 (SEPTEMBER 1996), pp. 504-505Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27969865 .

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equals 363 "between" days to be divided into the summer span and the winter span. An impossible pair of birthdays must have 180 or more days in both the summer span and the winter span. Conse quently, four summer-span

winter-span combinations exist for impossible pairs:

180 summer 183 winter 181 summer 182 winter 182 summer 181 winter 183 summer 180 winter

Specifically, assume that Larry was born on 1 April. If Michelle was born on 1 October, then the "between" summer span would be 29 (days in April) + the foregoing 153 = 182, and the "between" winter span would contain 30 (days in October) + the foregoing 151 = 181, hence an impossible pair. If Michelle was born on 2 October, then the BSS (between summer span) = 29 + 153 + 1 = 183 and the BWS (between winter span) is 29 October days + the foregoing 151 = 180; hence this pair, too, is impossible. Therefore, when Larry's birthday is 1 April, two impossible pairs are

generated. Assume that Larry is born on

2 April. If Michelle is born on 1 October, the BSS = 28 + 153 = 181 and the BWS = 30 (October) + 151 + 1 (April) = 182. Hence, an

impossible pair occurs. If Michelle is born on 2 October, then BSS = 28 + 153 + 1 = 182 and BWS = 29 + 151 + 1 = 181; an impossible pair. If Michelle is born 3 October, the BSS = 28 + 153 + 2 = 183 and BWS =180. Therefore, when Larry's birthday is 2 April, three impossible pairs occur.

If Larry is born on any other day in April, or April 3-30, then for each of these dates will be four "impossible" dates for Michelle in October. This result is perhaps surprising, so I will demonstrate for three April dates. If Larry is born on 3 April, then Michelle could be born on 1 October (BSS = 180 and BWS =

183) or 2 October (BSS = 181 and BWS = 182) or 3 October (BSS = 182 and BWS = 181) or 4 October (BSS = 183 and BWS =180).

If Larry is born on 17 April, then Michelle could be born on 15 October (BSS = 180 and BWS =

183) or 16 October (BSS = 181 and BWS = 182) or 17 October (BSS = 182 and BWS = 181) or 18 October (BSS =183 and BWS = 180).

If Larry is born on 30 April, then Michelle could be born on 28 October (BSS = 180 and BWS =

183) or 16 October (BSS = 181 and BWS = 182) or 17 October (BSS = 182 and BWS = 181) or 18 October (BSS = 183 and BWS = 180).

The following chart summa rizes these results:

Larry's No. of Impossible Birthday Dates for Michelle

1 April 2 2 April 3 3-30 April 4

Consequently, we have 112 + 3 + 2 = 117 impossible pairs for which we could not say that these two people had birthdays within 180 days of each other. We have 930 - 117 = 813 pairs of dates for which we can say that these two people were born with in 180 days of each other.

Finally the probability that Larry and Michelle have birthdays within 180 days of each other is 813/930. The solution offered seems to focus only on the "between" summer span and

ignores the between" winter span. Thank you very much for the

monthly Calendar and questions. We enjoy this feature of the jour nal very much.

Dave Stouffer Central High School Saint Joseph, MO 64501

Problem 2, April 1996 With all due respect to Blaise Pascal, it is unquestionably the case that he did not "invent" what is commonly referred to as Pascal's triangle. A brief inspec tion can locate published versions of the triangle in the Szu Yuen Yu Chien by Chu Shih Chieh (1303) and the Rechnung by Peter Apian (1527), both certainly predating Pascal's use of the device. It was also used by the Arab mathematician Jamshid al-Kashi in the fifteenth century and by Tartaglia, Stifel, and Stevin in Europe in the sixteenth century, among others. See

Boyer's A History of Mathmatics, Joseph's The Crest of the Peacock,

or Kline's Mathematical Thought from Ancient to Modern Times.

Donald W. Smith Smithd@aa.edu Albuquerque Academy Albuquerque, NM 87109

Wong responds: Alas and alack! But of course you are

right! My word choice was most unfortunate. "Pascal's triangle" is

named as such because of the extensive writings that Blaise made concerning the triangle and its properties. Please accept my apologies! J? suis desole?!

Problem 7, April 1996 Problem 7 in the April 1996 Cal endar defines a lattice point as a point (x, y) with integral coordi nates and shows that twelve lat tice points lie on the graph of x2 +

y2 = iV, where = 25. Consider the following generalization.

Problem: How many lattice points lie on the graph of x2 +

y2 = N, where is a positive integer? Solution: Let equal the num

ber of divisors of N, including 1, that have the form (4z + 1) where

is a nonnegative integer. Let m equal the number of

divisors of iV that have the form (4e - 1) where is a positive integer.

Then L, the number of lattice points, equals 4(p

- m), which is

never negative because > m for any A .

Therefore N= 50 oriV=100, for example, also would produce twelve lattice points as = 25 does. Divisors of that are even are ignored. The lattice points for iV = 50 are (?1, ?7), (?7, ?1), and (?5, ?5). The lattice points for iV = 100 are (?6, ?8), (?8, ?6), (0, ?10), and (?10, 0) Consider the follow ing three examples:

If = 9, two divisors, 1 and 9, are of the form (4e + 1) and one divisor, 3, is of the form (4e -1). L = 4(2-1) = 4. The four lat tice points are (0, ?3) and (?3,0). If = 15, two divisors, 1 and 5, are of the form (4e + 1) and two divisors, 3 and 15, are of the form (4z -1). L = 4(2 - 2), or 0. No lattice points exist. If = 25, all three divisors, 1, 5, and 25, are of the form (4z + 1). L = 4(3 - 0) = 12. The

twelve lattice points are as shown in the calendar problem.

Table 1 shows the smallest positive integer such that the graphs of x2 + y2 = will have L lattice points, where L = in and

is a nonnegative integer less than or equal to 16. Students may wish to list the divisors of N, verify and extend the table, and discover about it such generaliza tions as the following: .

Is it always true that is odd if and only if the smallest is a

square? [Yes, because an inte ger has an odd number of divi sors if and only if it is the square of an integer]

When = 6,10, and 14, small est iV=52 13, 54xl3, and 56 13, respectively. Is it always true that smallest = 5a 13 when = 2(a + 1) and a is a positive even integer? [No, because when = 18, smallest N iV = 52xl32xl7 = 71 825]

When = 8,12,16, and 20, smallestiV=5xl3xl7,52x 13 17, 53 13 17, and 54 13 17, respectively. Is it always true that smallest = 5b 13 17 when = 4(6 + 1) and b is a positive integer? [No, because when = 24, smallest N=52x 13x17x29 = 160 225]

When > 1, letting =5 A

will always produce L = in lattice points. However, it

may not be the smallest to do so. When > 2, is it always true that is prime if and only if smallest = ?""1? Note that if smallest iV = 1 = 51"1 =

5?, then = 1, which is not prime. For that reason, this "if and only if" question excludes = 1 because it would create an instant counterexample. This question is left for the reader to investi gate and, if necessary, find a

counterexample.

On page 96 of his book Recre ations in the Theory of Numbers (New York: Dover Publications, 1964), Albert H. Beiler states that "one of the mathematical luminaries of the early nineteenth century" was C. G. J. Jacobi. On page 142 of that book, Beiler states that "Jacobi gave an elegant rule," which is the expression for L in terms of and m used in this let ter. However, I know neither an

504 THE MATHEMATICS TEACHER

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alternative published source for this rule nor any source for its proof. Table 1 and any incorrect or incorrect generalizations about it are my creation.

Robert H. Becker 526 Harding Avenue Shillington, PA 19607-2802

Grant McLaughlin responds: The generalization offered by Becker is impressive. The inves tigative nature of his presentation is particularly appealing. The tabular summary nicely substan tiates many of his own insights into the problem. Meanwhile, an invitation is extended to others who may wish to contribute their own ideas.

I am grateful for Becker's efforts to share these reflections with the readership. His efforts exemplify how a seemingly basic question is usually but one

aspect of a much more complex picture.

Problem 19, April 1996

Given that K2 - 3K + 5 = 0, determine the value of K4 -

6K3+9K2-7.

This Calendar problem presents a good chance to practice division of polynomials.

K2 -3K-5

K2 -3K+5 K4 -6K3 +9K2+OK+7

K4 -3K3 +5K2

-3K3 +4K2+OK-7

-3K3 +9K2-15K

-5K2 +15K-7

-5K2 +15K-25

+18

Since

K2 -3K+ 5 =0

and

(K2-3K+ 5)(K2-3K-5)+ 18 = K4 -6K3 + 9K2+7,

: ' 18=K4-6K3+9K2+7. David Pease Summit High School Summit, NJ 07901

Grant McLaughlin responds: Indeed the problem does lend

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Vol. 89, No. 6 e September 1996 505

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