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Answers and Analysis of Compulsory Two Exercises in Senior High School Mathematics
If when 3n+ 1 is a complete square number, n+ 1 can be expressed as the sum of k complete squares, then the minimum value of k is ().

Solution: It is known that 3n+ 1 is a complete square number, so let's set 3n+ 1=a2.

Obviously a2 is not a multiple of 3, so A = 3x 1.

Therefore, 3n+ 1 = A2 = 9X2 6x+ 1, n = 3x2x2,

That is n+1= 2x2+(x1) 2 = x2+x2+(x1) 2,

That is, n+ 1 is written as the sum of squares of x, x, X 1.

That is to say, expressed by the sum of three complete squares,

So k = 3.

So choose C.

Do it yourself according to the principle and the method will tell you.