Q2 ALGEBRA

Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?

a) The minimum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m + 1

b) The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1

c) The maximum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m +1

d) The maximum possible value of a2 + b2 +c2 + d2 is 4m2 + 2m + 1

(a + b + + d)2 = (4m + 1)2 

Thus, a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd) = 16m2 + 8m + 1

a2 + b2 + c2 + d2 will have the minimum value if (ab + ac + ad + be + bd + cd) is the maximum.

This is possible if a=b=c=d= (m + 0.25) ... since a +b+c+d= 4m+ 1

In that case 2(ab + ac + ad + bc + bd + cd) = 12(m + 0.25)2 = 12m2 + 6m + 0.75

Thus, the minimum value of a2 + b2 + c2 + d2 = (16m2 + 8m + 1)— 2(ab + ac + ad + bc + bd + cd) = (16m2 + 8m + 1)— (12m2 + 6m + 0.75) = 4m2 + 2m + 0.25

Since it is an integer, the actual minimum value = 4m2 + 2m + 1

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