The following term: $$\sqrt{2017^2-2018^2+2019^2}$$ is the same as this term: $$\sqrt{2018^2+2}$$ how can one show without the need of a calculator that these are the same?

The original question was the following: How long is the distance between point A and D - the diagonals are orthogonal and the other distances as labeled

I then called the point where the diagonals meet $M$ and with the help of the Pythagoras' theorem I came up with the following equations:

$$ AM^2 + BM^2 = 2017^2 $$ $$ AM^2 + CM^2 = 2018^2 $$ $$ CM^2 + DM^2 = 2019^2 $$ $$ DM^2 + AM^2 = AD^2 $$

By rearrange the first three equations I had the following term:

$$AD^2 = 2017^2-2018^2+2019^2 $$ as the solution given was: $$ AD = \sqrt{2018^2+2}$$ I was curious how to tell it was the same

Are there any other "nicer" ways to find this solution?

  • $\begingroup$ How were you planning to do it with a calculator? $\endgroup$ – Rob Arthan Aug 12 '17 at 22:24
  • $\begingroup$ What are these square roots doing here? $\endgroup$ – José Carlos Santos Aug 12 '17 at 22:25
  • $\begingroup$ the root comes from the actual problem - but it's not important $\endgroup$ – oneguy Aug 12 '17 at 22:35
  • 1
    $\begingroup$ @oneguy the root comes from the actual problem - but it's not important It's still important in the sense that the actual problem might well have a simpler or more direct solution without going through those square roots. See What is the XY problem?. $\endgroup$ – dxiv Aug 12 '17 at 23:29
  • $\begingroup$ oh, I'm sorry - I'll change it $\endgroup$ – oneguy Aug 13 '17 at 9:54



You deduce that $2017^2-2018^2+2019^2=2018^2-2\cdot2018+1-2018^2+2018^2+2\cdot2018+1=2018^2+2$


We know that the difference between square numbers increases by $2$ each time eg. because the sum of the first $n$ odd numbers is $n^2$.

Thus $(2019^2-2018^2)=(2018^2-2017^2)+2$ and thus $2019^2-2018^2+2017^2=2018^2+2$ as required


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