Let a,b,c,d,m,n∈Z+ such that a2+b2+c2+d2=1989, a+b+c+d=m2, and the largest of a,b,c,d is n2. Determine, with proof, the values of m and n.
Problem
Source: IMO Shortlist 1989, Problem 15, ILL 50
Tags: number theory, system of equations, Diophantine equation, Additive Number Theory, IMO Shortlist
joh
18.09.2008 18:50
m=9,n=6.
By Power Mean Inequality we have
a+b+c+d≤2√1989<90.
Since a+b+c+d and a2+b2+c2+d2 have the same parity, the possible values of m2 are 1,9,25,49,81. But m4=(a+b+c+d)2>a2+b2+c2+d2=1989, so m2 is either 49 or 81.
Assume that m2=49 and max. We have a + b + c = 49 - d and a^2 + b^2 + c^2 = 1989 - d^2. Since (a + b + c)^2 > a^2 + b^2 + c^2, we have
(49 - d)^2 > 1989 - d^2,
i.e.
d^2 - 49d + 206 > 0.
This gives d\ge45 or d\le4. If d\ge45, then d^2\ge2025 > 1989 = a^2 + b^2 + c^2 + d^2, which is impossible. If d\le4, then a + b + c + d\le16, which is impossible too.
Hence m^2 = 81. We have n^2\ge\frac {81}4\ge20 and n^2 < \sqrt {1989}\le44. So n^2 is either 25 or 36. Assume that n^2 = 25. Note that 25 - a,25 - b,25 - c are all nonnegative. We have (25 - a) + (25 - b) + (25 - c) = 19 and (25 - a)^2 + (25 - b)^2 + (25 - c)^2 = 439. This is impossible since 19^2 < 439. So n^2 = 36.
We can see easily that m=9,n=6 is possible. Indeed, a=12,b=15,c=18 satisfies the given equations.
diks94
18.01.2011 18:24
According to me the answer is m=9, n=6
As \frac{a^2+b^2+c^2+d^2}{4} > (\frac{a+b+c+d}{4})^2
By Weighted mean Ineaquality
putting values of a+b+c+d = m^2
and that of a^2 + b^2 + c^2 + d^2 =1989
we get m^4 <1989 * 4
m<9.4
as a,b,c,d,are Z+
m=9
max
for m=9
a+b+c+d= 81 and a^2 + b^2 + c^2 + d^2 =1989
let max of a,b,c,d is a
then b+c+d=81-a and b^2 + c^2 + d^2=1989 -a^2
as (b+c+d)^2 >b^2 + c^2 + d^2
solving this equation we get the value off n as 6