a, b, c, d, m, n are positive integers such that a2 + b2 + c2 + d2 = 1989, a + b + c + d = m2 and the largest of a, b, c, d is n2. Find m and n.
Solution
Answer: m = 9, n = 6.
74 = 2401 > 1989, so n < 7. We have 42 = 16 and 4 x 162 = 1024 < 1989, so n > 4. Hence n = 5 or 6. Assume a ≥ b ≥ c ≥ d, so a = n2.
k has the same parity as k2. But a2 + b2 + c2 + d2 = 1989 is odd, so a + b + c + d = m2 is odd. If n = 36, then b2 + c2 + d2 = 693, so by Cauchy, (b + c + d)2 ≤ 3.693. Hence b + c + d ≤ 45. Hence 36 < m2 ≤ 81 and m2 is odd, so m = 7 or 9. Similarly, if n = 25, then b2 + c2 + d2 = 1364 and so by Cauchy b + c + d < 64. Hence again m = 7 or 9.
Suppose first that m = 7. Then (49 - a)2 = (b + c + d)2 > b2 + c2 + d2 = 1989 - a2, so a2 - 49a + 206 > 0. Put f(x) = x2 - 49x + 206. We can easily check that f(4) = f(45) = 26, f(5) = f(44) = -14, so f(x) < 0 for 5 ≤ x ≤ 44. But we know that a must be 25 or 36, so there are no solutions with m = 7. Hence m = 9.
It is easy to check that we do have a solution for m = 9, n = 6: 12 + 15 + 18 + 36 = 92 and 122 + 152 + 182 + 362 = 1989.
Finally, suppose m = 9, n = 5. Then b + c + d = 56, b2 + c2 + d2 = 1364. Put b = 25 - B, c = 25 - C, d = 25 - D, where B, C, D are non-negative. So B + C + D = 19, B2 + C2 + D2 = 1364 - 1875 + 50.19 = 439. But now B2 + C2 + D2 > (B + C + D)2. Contradiction. So there are no solutions with m = 9, n = 5.
© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02