For a positive integer , define  to be the minimum value of the sum

\(\sum_{k = 1}^n \sqrt{(2k-1)^2 + (a_k)^2}, \)
where  are positive real numbers whose sum is 17. There is a unique positive integer  for which  is also an integer. Find this n.

pleaz halp :)

 Jun 24, 2024

Observe that the sum and apply it to the Cartesian plane. We can see that it is basically calculating the distance between the origin and n points.


Therefore, we can assume every (2k−1)2+ak2​​ to be a vector from the origin to point (2k−1,ak​).


Now we can take the sum inside the vector space and get ((1+3+5+…+(2n−1))2+(a12​+a22​+…+an2​))1/2 (by the Triangle Inequality, this is always less than or equal to the original sum). Note that 1+3+…+(2n−1)=n2.


We are given that a12​+a22​+…+an2​=17, so Sn​≥n4+17​. We can achieve equality by taking a1​=a2​=…=an​=n17​​.


In this case, all the points would be on a circle centered at the origin with radius n2+n17​​=nn3+17​​. Thus, Sn​=n4+17​ if and only if n3+17 is a perfect square.


We can now check small values of n. We see that n=2 does not work, n=3 does not work, but n=4​ does work, since 43+17=89 is a perfect square.


e can prove that there are no other solutions by Method of Descent. Suppose there exists another solution n′>4. Then (n′)3+17>43+17=89, so (n′)3>89.


By Integer Root Theorem, the only possible integer roots of (n′)3−89 are 1 and −1. We can quickly check that neither 1 nor −1 is a root.


This forces (n′)3−89 to be prime. However, this is impossible because (n′−4)(n′2+4n+22) divides (n′)3−89, and both factors on the right are greater than 1.


Thus, n=4 is the unique solution.

 Jun 25, 2024

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