 31.
 Find all continuous functions f: R ®R such that

 32.
 Let n ³ 3 be a natural number. Prove that

 33.
 Let n ³ 3 be a positive integer. Are there n positive integers a_{1}, a_{2}, ¼, a_{n} not all the same such that for each i with 3 £ i £ n we have

 34.
 Find all integers p for which there are rational numbers a and b such that the polynomial x^{5}  px  1 has at least one root in common with the polynomial x^{2}  ax  b.
 35.
 Let P be an arbitrary interior point of an equilateral triangle ABC. Prove that

 36.
 Let n be a positive integer and x > 0. Prove that

 31.
 First solution. Setting (x, y) = (t, 0) yields f(t + f(0)) = f(t) for all real t. Setting (x, y) = (0, t) yields f(f(t)) = f(0) + t for all real t. Hence f(f(f(t))) = f(t) for all real t, i.e., f(f(z)) = z for each z in the image of f. Let (x, y) = (f(t), f(0)). Then

 Taking (x, y) = (u, f(v)) yields

 31.
 Second solution. Taking (x, y) = (0, 0), we find that

 31.
 Third solution. Taking (x, y) = (0, 0) yields f(f(0)) = f(0), whence f(f(f(0))) = f(f(0)) = f(0). Taking (x, y) = (0, f(0)) yields f(f(f(0))) = 2f(0). Hence 2f(0) = f(0) so that f(0) = 0. Taking x = 0 yields f(f(y)) = y for each y. We can complete the solution as in the Second Solution.
 32.
 First solution. Let N = n^{nnn}  n^{nn}. Since 1989 = 3^{2} ·13 ·17,

 (i) x^{u} º 0 (mod 9) whenever u ³ 2 and x º 0 (mod 3).
 (ii) x^{6} º 1 (mod 9) whenever x \not º 0 (mod 3).
 (iii) x^{u} º 0 (mod 13) whenever x º 0 (mod 13).
 (iv) x^{12} º 1 (mod 13) whenever x \not º 0 (mod 13), by Fermat's Little Theorem.
 (v) x^{u} º 0 (mod 17) whenever x º 0 (mod 17).
 (vi) x^{16} º 1 (mod 17) whenever x \not º 0 (mod 17), by FLT.
 (vii) x^{4} º 1 (mod 16) whenever x = 2y + 1 is odd. (For, (2y+1)^{4} = 16y^{3}(y+2) + 8y(3y+1) + 1 º 1 (mod 16).)
 Note that

 Modulo 17. If n º 0 (mod 17), then n^{nn} º 0, and so N º 0 (mod 17).
 If n is even, n ³ 4, then n^{n} º 0 (mod 16), so that

 Suppose that n is odd. Then n^{n} º n (mod 4)





 Modulo 13. If n º 0 (mod 13), then n^{nn} º 0 and N º 0 (mod 13).
 Suppose that n is even. Then n^{n} º 0 (mod 4), so that n^{nn}  n^{n} º 0 (mod 4). Suppose that n is odd. Then n^{nn  n}  1 º 0 (mod 16) and so n^{nn}  n^{n} º 0 (mod 4).
 If n º 0 (mod 3), then n^{n} º 0 so n^{n}(n^{nn  n}  1) º 0 (mod 3). If n º 1 (mod 3), then n^{nn  n} º 1 so n^{n}(n^{nnn}  1) º 0 (mod 3). If n º 2 (mod 3), then, as n^{n}  n is always even, n^{nnn} º 1 so n^{n}(n^{nnn}  1) º 0 (mod 3). Hence, for all n, n^{nn}  n^{n} º 0 (mod 3).
 It follows that n^{nn}  n^{n} º 0 (mod 12) for all values of n. Hence, when n is not a multiple of 13, n^{(nnn  n)} º 1 so N º 0 (mod 13).
 Modulo 9. If n º 0 (mod 3), then n^{nn} º 0 (mod 9), so N º 0 (mod 9). Let n \not º 0 (mod 9). Since n^{nn}  n^{n} is divisible by 12, it is divisible by 6, and so n^{(nnn  nn)} º 1 and N º 0 (mod 9). Hence N º 0 (mod 9) for all n.
 The required result follows.
 33.
 First solution. Letting b_{i} = (a_{i}, S_{i}), we find that




 33.
 Second solution. Let a_{i} = 1, a_{2} = 2 and a_{i} = 3 ·2^{i3} for i ³ 3. Then


 33.
 Third solution. [K. Purbhoo] Choose a_{1} at will, and let a_{i} = S_{i1} for i ³ 2. Then S_{i} = S_{i1} + a_{i} = 2a_{i}, a_{i} + S_{i} = 3a_{i}, (a_{i}, S_{i}) = a_{i} and [a_{i}, S_{i}] = 2a_{i} for i ³ 2.
 33.
 Fourth solution. [K. Yeats] Let a_{1} = 1, a_{2} = 3 and a_{n} = 2^{n1} for n ³ 3. Then S_{n} = 2^{n} for n ³ 2 and, for each i with 3 £ i £ n, (a_{i}, S_{i}) = 2^{i1} and [a_{i}, S_{i}] = 2^{i}.
 34.
 First solution. A common rational root must be 1 or 1, in which case the respective values of p are 0 and 2. These roots are shared with x^{2}  1 (a = 0, b = 1). Suppose that there is a common nonrational root r. Then r^{2} = ar + b so that






 34.
 Second solution. The rational root case can be handled as in the first solution. By long division, we find that


 35.
 First solution. The result is clear if P is on the bisector of the angle at A, since both sides of the inequality are 0.
 Wolog, let P be closer to AB than AC, and let Q be the image of P under reflection in the bisector of the angle A. Then


 Produce PQ to meet AB in R and AC in S. Consider the reflection Â with axis RS. The circumcircle C of DARS is carried to a circle C¢ with chord RS. Since ÐRCS < 60^{°} = ÐRAS and the angle subtended at the major arc of C¢ by RS is 60^{°}, the point C must lie outside of C¢. The circumcircle D of DAPQ is carried by Â to a circle D¢ with chord PQ. Since D is contained in C, D¢ must be contained in C¢, so C must lie outside of D¢. Hence ÐPQC must be less than the angle subtended at the major arc of D¢ by PQ, and this angle is equal to ÐPAQ. The result follows.
 36.
 First solution. By the ArithmeticGeometric Means Inequality, we have that


 36.
 Second solution (by calculus). Let


 36.
 Third solution. Let u = nx  1 so that x = (1 + u)/n. Then




