Each of the numbers1, 5 = 1 + 4, 12 = 1 + 4 + 7, 22 = 1 + 4 + 7 + 10, ...represents the

Each of the numbers 1=1, 3 = 1+2, 6 = 1+2 + 3, 10 = 1+2 + 3 + 4, ... represents the number of dots that can be arranged evenly in an equilateral triangle: This led the ancient Greeks to call a number triangular if it is the sum of consecutive integers, beginning with 1. Prove the following facts concerning triangular numbers: (a) A number is triangular if and only if it is of the form n(n + 1)/2 for some n 2: 1. (Pythagoras, circa 550 B.C.) (b) The integer n is a triangular number if and only if 8n + 1 is a perfect square. (Plutarch, circa 100 AD.) ( c) The sum of any two consecutive triangular numbers is a perfect square. (Nicomachus, circa 100 AD.) (d) If n is a triangular number, then so are 9n + 1, 25n + 3, and 49n + 6. (Euler, 1775)

If tn denotes the nth triangular number, prove that in terms of the binomial coefficients, _ n + 1) fn - \ 2 n 2: 1

Derive the following formula for the sum of triangular numbers, attributed to the Hindu mathematician Aryabhata (circa 500 A.o.): n(n + l)(n + 2) t1 + t2 + t3 + + tn = 6 n 2: 1 [Hint: Group the terms on the left-hand side in pairs, noting the identity tk-I + tk = k2 .]

Prove that the square of any odd multiple of 3 is the difference of two triangular numbers; specifically, that 9(2n + 1)2 = t9n+4 - t3n+l

In the sequence of triangular numbers, find the following: (a) Two triangular numbers whose sum and difference are also triangular numbers. (b) Three successive triangular numbers whose product is a perfect square. ( c) Three successive triangular numbers whose sum is a perfect square.

(a) If the triangular number tn is a perfect square, prove that t4n(n+l) is also a square. (b) Use part (a) to find three examples of squares that are also triangular numbers.

Show that the difference between the squares of two consecutive triangular numbers is always a cube

Prove that the sum of the reciprocals of the first n triangular numbers is less than 2; that IS, 1 1 1 1 1 - + - + - + - + + - <2 1 3 6 10 [Hint: Observe that n(nl) = 2( - nl ).]

(a) Establish the identity tx = ty + tz, where n(n + 3) x = + 1 2 y=n +l z= n(n + 3) 2 and n 2:: 1, thereby proving that there are infinitely many triangular numbers that are the sum of two other such numbers. (b) Find three examples of triangular numbers that are sums of two other triangular numbers.

Each of the numbers 1, 5 = 1 + 4, 12 = 1 + 4 + 7, 22 = 1 + 4 + 7 + 10, ... represents the number of dots that can be arranged evenly in a pentagon: The ancient Greeks called these pentagonal numbers. If Pn denotes the nth pentagonal number, where P1 = 1 and Pn = Pn-1 + (3n - 2) for n 2:: 2, prove that n(3n - 1) Pn = 2 n 2:: 1

For n 2:: 2, verify the following relations between the pentagonal, square, and triangular numbers: (a) Pn = tn-l + n2 (b) Pn = 3tn-l + n = 2tn-l + tn

Prove that, for a positive integer n and any integer a, gcd(a, a+ n) divides n; hence, gcd(a, a+ 1) = 1.

Given integers a and b, prove the following: (a) There exist integers x and y for which c =ax+ by if and only if gcd(a, b) I c. (b) If there exist integers x and y for which ax+ by= gcd(a, b), then gcd(x, y) = 1.

For any integer a, show the following: (a) gcd(2a + 1, 9a + 4) = 1. (b) gcd(5a + 2, ?a+ 3) = 1. (c) If a is odd, then gcd(3a, 3a + 2) = 1.

If a and b are integers, not both of which are zero, prove that gcd(2a - 3b, 4a - 5b) divides b; hence, gcd(2a + 3, 4a + 5) = 1.

Given an odd integer a, establish that a 2 +(a+ 2)2 +(a+ 4)2 + 1 is divisible by 12.

Prove that the expression (3n)!/(3!)n is an integer for all n =:::: 0.

Prove: The product of any three consecutive integers is divisible by 6; the product of any four consecutive integers is divisible by 24; the product of any five consecutive integers is divisible by 120. [Hint: See Corollary 2 to Theorem 2.4.]

Establish each of the assertions below: (a) If a is an arbitrary integer, then 61 a(a2 + 11). (b) If a is an odd integer, then 241 a(a2 - 1). [Hint: The square of an odd integer is of the form 8k + 1.] (c) If a and bare odd integers, then 81 (a2 - b2 ). (d) If a is an integer not divisible by 2 or 3, then 241(a 2+23). (e) If a is an arbitrary integer, then 360 I a2 (a2 - l)(a2 - 4).

Confirm the following properties of the greatest common divisor: (a) If gcd(a, b) = 1, and gcd(a, c) = 1, then gcd(a, be)= 1. [Hint: Because 1 =ax+ by= au+ cv for some x, y, u, v, 1 =(ax+ by)(au + cv) = a(aux + cvx + byu) + bc(yv).] (b) If gcd(a, b) = 1, and c I a, then gcd(b, c) = 1. (c) If gcd(a, b) = 1, then gcd(ac, b) = gcd(c, b). (d) If gcd(a, b) = 1, and c I a+ b, then gcd(a, c) = gcd(b, c) = 1. [Hint: Let d = gcd(a, c). Then d I a, d I c implies that d I (a+ b) - a, or d I b.] (e) If gcd(a, b) = 1, d I ac, and d I be, then d I c. (f) If gcd(a, b) = 1, then gcd(a 2 , b2 ) = 1. [Hint: First show that gcd(a, b 2 ) = gcd(a 2 , b) = l.]

(a) Prove that if d I n, then 2 d - 1 12n - 1. [Hint: Use the identity x k - 1 = (x - l)(x k - l + x k - 2 + + x + l).] (b) Verify that 2 35 - 1 is divisible by 31 and 127.

Let tn denote the nth triangular number. For what values of n does tn divide the sum ti + t1 + + tn? [Hint: See Problem l(c), Section 1.1.]

If a I be, show that a I gcd(a, b) gcd(a, c)