This was a curious math problem I found online, which appeared in some Romanian Math Olympiad. I was able to do it with some high school mathematics, and I think the solution was elegant and wanted to share it with you. But first, I want you to try it out. So here it is :

Prove that the integer part of the sum of squares of 5 up to 13 is 26, without using the exact values of the square roots.

$$ \sqrt5 + \sqrt6 + \sqrt7 + \dots + \sqrt12 + \sqrt13 \, = \, 26.\\ $$

(Bonus points if you can produce the range for the decimal)

I have given a hint to get you started (if you need it):

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Hint 1: Start thinking about the integral ranges of each square root. 9 has a square root of 3, lower numbers have it between 2 and 3, higher numbers have it between 3 and 4. By this, you already have it between 23 and 31. Now think of any pattern that you might find useful !!

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Solution:

Let’s get started now !! Let’s denote the sum by S, and by the hint, we reduced the range by a bit:

$$ \sqrt5 + \sqrt6 + \sqrt7 + \sqrt8+ \sqrt9+ \sqrt10+ \sqrt11 + \sqrt12 + \sqrt13 \, = S $$

We get that there is a 9 with square root 3, with 4 numbers above and 4 below to be added.

We also noted that lower numbers have it between 2 and 3, higher numbers have it between 3 and 4. Thus, it gives :

$$ 8 <\sqrt5 + \sqrt6 + \sqrt7 + \sqrt8 < 12 $$