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As the title says, this is a problem for a mathematics contest, no calculators allowed.

Contestants should solve this problem within a few minutes or so, therefore brute-force is a no-go.

I've tried substituting the constants:

$$\sqrt{a^2+(ab)^2+b^2}$$

I have the idea of using $$(a+b)^2=a^2+b^2+2ab$$

But I still can't see a way to make the calculation simple, please help!

Edit: Some people are saying that "a few minutes" is too short. In a math contest, if you finish all the other problems fast, you may have extra time to solve this particular problem, so don't be afraid to share your thoughts, even if it may take 5 or 10 minutes.

Cyh1368
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    Take $b=a+1$ in your calculations. After expanding and regrouping, you will find a degree 4 polynomial which can be factored to the square of an expression in "a". So finally, you will have to calculate $2016^2$ and add it to itself and 1. – MasB Feb 28 '22 at 13:23
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    A minute?? Maybe I get this in $10$ minutes, but $1$ minute is too fast... – Adam Rubinson Feb 28 '22 at 13:27
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    A method is to calculate it for $a = 1, 2, 3...$ and finally find that $a^2 + (a+1)^2 + a^2 (a+1)^2 = (a^2+a + 1)^2$. But one minute is short... – Damien Feb 28 '22 at 13:32
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    One minute is ridicoulous for this exercise, although surely some freaks will be able to solve it within this time. Do you have to write down the solution ? Then, you have only about $30$ seconds to think ! – Peter Feb 28 '22 at 13:39
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    The question doesn't make sense, as you will admit that one minute is too short. If you want to train for speed you have to know many formulas and tricks by heart. That you need to ask here for help means that you haven't started yet learning this by heart. – Dietrich Burde Feb 28 '22 at 13:39
  • By the way , pure brute force is hopeless anyway without a calculator considering the big numbers. So, there is no need to limit the time so drastically to see whether someone found the trick. Without it , noone will solve this within a reasonable amount of time. – Peter Feb 28 '22 at 13:44
  • @Peter No, getting the solution is enough. – Cyh1368 Feb 28 '22 at 14:06
  • If you let $a=10$ you get $\sqrt{100+12100+121}=\sqrt{12321}$ – Empy2 Feb 28 '22 at 14:19
  • the closest square is 2025, so if we take it as c, then 2016 is c-9, and 2017 is c-8, maybe that will reduce it to a 4th degree polynomial? – MonK Feb 28 '22 at 17:14

1 Answers1

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One possibility is to define the function

$$ f(a) = \sqrt {a^2 + (a+1)^2 + a^2(a+1)^2} $$

To calculate it for $a=0,1,2,3\dots$ and to analyze the result with the finite differences method.

$$ \begin{array}{c|rrrrrr} a & 0 & 1 & 2 & 3 & 4 \\ \hline f(a) & 1 & 3 & 7 & 13 & 21 \\ \Delta_1 & & 2 & 4 & 6 & 8 \\ \Delta_2 & & & 2 & 2 & 2 \\ \Delta_3 & & & & 0 & 0 \\ \end{array} $$

And then to guess that the result is a polynomial of order 2.

At the end, we get that

$$ a^2 + (a+1)^2 + a^2(a+1)^2 = (a^2 + a + 1)^2 $$

But effectively, one minute is short.

Damien
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