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Solution for May 2007

The Problem:

Which positive integers n satisfy


for all real numbers xn?

Correct responses:

We received correct submissions for this month's problem from the following people:

Martin Argerami (Regina)  

Normand LaLiberté (Ontario)

Christophe Brighi (France)


Matthew Lim (USA)

Qin Deng (Toronto)


John T. Robinson (USA)

Sébastien Dumortier (France)


K. Sengupta (India)

Stefan Gatachiu (Romania)


Bob Franz (USA)

Wolfgang Kais (Germany)   Xavier Hecquet (France)


The solution:

Only n = 1 and n = 2 satisfy the given inequality.
            Indeed, for n = 1 we have x – 1 < x, so that for x ≥ 1


as required.
            For n = 2 and x ≥ 2, start with (x – 2)2 ≥ 0, which implies x2≥ 4x – 4.  Taking the square root of both sides yields


as required.

            For n ≥ 3 the inequality fails for appropriate values of x.  In fact, it always fails for x = n + 1. We base our proof of the claim on two very simple inequalities:

  1. . for j running from 0 to n – 1, and
  2. .(To verify this inequality one can use a calculator, or note that since ., it follows that

Proof that .when x = n + 1:


            In addition to solving our problem, John Robinson investigated the behavior of the difference between x and the sum of the square roots, namely


As we saw above, for n ≥ 3 the function is negative when x = n+1; moreover, it is clearly positive for large x, say x > n2.  Robinson was able to show that f(x) switches from negative to positive around n2n/2.  More precisely, when n > 4 and x is an integer,


where floor brackets designate the greatest integer less than or equal to the number they enclose.  So, for example, setting n = 5 and x = 52 – (5 + 1)/2, we get




He managed to show this by replacing each square root by its second degree Taylor polynomial


Meanwhile, Matthew Lim wondered what sort of result holds when the square roots in our problem are replaced by kth roots.  He was able to show (by induction) that as with square roots, for n ≥ 2k – 1 and x = n + 1, we have


He was, however, unable to say anything about how the sum compares to x when n < 2k – 1, except in the case of our problem (where k = 2).




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