Mathematical Problems & Inquiry in Mathematics AME Tenth Anniversary Meeting May A/P Peter Pang Department of Mathematics and University Scholars Programme, NUS

Four Important Concepts Specificity Generality Specialization Generalization

DF37  Each card has a number on one side and a letter on the other.  Claim: “If a card has ‘D’ on one side, then it has a ‘3’ on the other.”  Which cards do you need to turn over to find out if this is true?

 You are a bouncer in a bar. You must make sure that there are no under-age (below 21) drinkers.  There are 4 customers (A—D) in the bar. You know what 2 of them are drinking and you know the age of the other 2. Customer A is drinking beer Customer B is drinking coke Customer C is 25 years old Customer D is 16 years old  Which of the 4 customers do you need to check to do your job?

What is a maths problem?  One (possibly the most?) important aspect of inquiry in mathematics is to find a problem  One important quality of a maths problem has to do with the notions of specificity and generality  Consider the following problems: If x = 2 and y = 4, show that x + y = 6 If x is even and y is even, show that x + y is even

 Define “even”  A number is even is it is two times a natural number; equivalently, an even number is divisible by 2, i.e., when divided by 2, the remainder is 0.  x is even if x = 2n for some natural number n  Let x = 2n and y = 2m where n and m are natural numbers  x + y = 2n + 2m = 2(n+m)  As n and m are natural numbers, so is n + m. This shows that x + y is even.

Can this be generalized?  If x is a multiple of 3 and y is a multiple of 3, then so is x + y.  If x and y are multiples of p, then so is x + y.  If x is odd and y is odd, is x + y odd?  The number x is odd if, when divided by 2, the remainder is 1. We denote this by x = 1 (mod 2).  If x = 1 (mod 2) and y = 1 (mod 2), is x + y = 1 (mod 2)?

 If x = 1 (mod 2) and y = 1 (mod 2), then x + y = (mod 2) = 2 (mod 2) = 0 (mod 2)  This means that x + y is even.  If x = p (mod r) and y = q (mod r), where p, q, r are natural numbers, then x + y = p + q (mod r)

Tension between Specificity and Generality  Generality is often accompanied by loss of context (i.e., abstractness)

DF373 C2516B D 21B D73D

Comparison with other disciplines  Literature  Science  Social science

A less trivial example  x 2 + y 2 = z 2 has an infinite number of positive integer solutions x = u 2 – v 2 y = 2uv z = u 2 + v 2  This result is believed to be due to Pythagoras  What about powers higher than 2?

Fermat’s Last Theorem  x n + y n = z n has no integer solution when n > 2  Observation #1 (specialize to prime powers):  It suffices to look at powers n that are prime  Suppose there is a solution (x, y, z) for n = p x q, where p is prime. Then x pq + y pq = z pq (x q ) p + (y q ) p = (z q ) p  Thus, (x q, y q, z q ) would be an integer solution for the power p.

 Very important note:  If you have a solution for the power pq, then you have a solution for the power p (and q)  However, if you have a solution for the power p, it does not mean that you have a solution for the power pq  (x q ) p + (y q ) p = (z q ) p

 Observation #2 (specialize to “primitive solutions”)  It suffices to look at solutions that are pairwise relatively prime, i.e., between any two of the three numbers x, y and z there are no common factors other than 1  For example, suppose x and y have a common factor of 2. Then, as x n + y n = z n, z will also have a factor of 2. Thus I can divide the equation through by the common factor 2.

 Observation #3 (generalize to rational solutions)  Instead of asking for integer solutions, the problem can be equivalently stated by asking for rational solutions x = a/b y = c/d z = e/f (a/b) p + (c/d) p = (e/f) p  Put the three fractions under a common denominator g (a’/g) p + (c’/g) p = (e’/g) p a’ p + c’ p = e’ p

A special case : n = 4  To show that x 4 + y 4 = z 4 has no integer solution  Strategy: Proof by contradiction  Suppose there were a solution. Will show that this supposition will lead to a logical contradiction, i.e., something will go wrong.  As a result, the supposition cannot be correct, and hence its opposite is correct.

 The solution for n = 4 uses the very interesting idea of “infinite descent”  Suppose there were a solution (x, y, z), i.e., x 4 + y 4 = z 4  Write z 2 = w. Then x 4 + y 4 = w 2 or (x 2 ) 2 + (y 2 ) 2 = w 2  By Pythagoras x 2 = u 2 – v 2, y 2 = 2uv, w = u 2 + v 2  From this, we get x 2 + u 2 = v 2

 Again by Pythagoras, x = s 2 – t 2 u = 2stv = s 2 + t 2  Recalling that y = 2uv, we have y 2 = 2(2st)(s 2 + t 2 )  and hence (y/2) 2 = st(s 2 + t 2 )  Note that s, t and s 2 + t 2 are relatively prime.  As their product is a perfect square, so must each individual factor (s, t and s 2 + t 2 ).

 This means that s = x 1 2 t = y 1 2 s 2 + t 2 = w 1 2 and hence x y 1 4 = w 1 2  Finally, note that x 1 < xy 1 < yw 1 < w  This leads to infinite descent, which is not possible as we are dealing with positive integers.

Conclusions