History of Math

History of Math

Answer 1

Suppose that we start with a primitive Pythagorean triple (a,b,c). If any two of a,b,c shared a common divisor d, then, using the equation

a² + b² = c²,

We could see that d² would have to divide the square of the remaining one, so that d would have to divide all three. This would contradict the assumption that our triple was primitive. Thus no two of a, b, and c have a common divisor greater than 1.

Now notice that not all of a, b, and c can be odd, because if a = 2x + 1, b = 2y + 1, and c = 2z + 1, then the equation would be

(2x+1)² + (2y+1)² = (2z+1)²,

4x² + 4x + 1 + 4y² + 4y + 1 = 4z² + 4z + 1,

4(x² + x + y² + y - z² - z + 1) = 3.

This implies that 4 is a divisor of 3, which is false. Now suppose that c were even, and a and b odd, so a = 2x + 1, b = 2y + 1, and c = 2z. Now the equation would be

(2x+1)² + (2y+1)² = (2z)²,

4x² + 4x + 1 + 4y² + 4y + 1 = 4z²,

4(x² + x + y² + y - z² + 1) = 2.

This implies that 4 is a divisor of 2, which is also false. Thus either a or b must be even, and the other two odd. Let's say it is b that is even, with a and c odd. (If not, switch the meanings of a and b in what follows below.)

Now rewrite the equation in the form

b² = c² - a²,

(b/2)² = ([c - a]/2)([c + a]/2).

Notice that since b is even and a and c are odd, b/2, (c - a)/2, and (c + a)/2 are whole numbers, and all are positive. Now we claim that the greatest common divisor of (c - a)/2 and (c + a)/2 is 1. If d is any common divisor of these, then d would divide their sum, c, and their difference, a. We know, however, that the greatest common divisor of a and c is 1, so d must divide 1, so d = 1.

Answer 2 (a)

Students often encounter formulas for sums of powers of the first n positive integers as examples of statements that can be proved using the Principle of Mathematical Induction and, perhaps less often nowadays, in Riemann sums during an introduction to definite integration.  In either situation, they usually see only the first three such sum formulas,

 

1+2+3+  +n= 2n(n+1)

 

12 +22 +32 +  +n2 = 6n(n+1)(2n+1)

and

13 +23 +33 +  +n3 =4n2 (n+1)2

for any positive integer n. 

 

Formulas for sums of integer powers were first given in generalizable form in the West by Thomas Harriot (c. 1560-1621) of England.  At about the same time, Johann Faulhaber (1580-1635) of Germany gave formulas for these sums up to the 17th power, far higher than anyone before him, but did not make clear how to generalize ...