# simplify a numerical expression without calculator

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?

• How were you planning to do it with a calculator? – Rob Arthan Aug 12 '17 at 22:24
• What are these square roots doing here? – José Carlos Santos Aug 12 '17 at 22:25
• the root comes from the actual problem - but it's not important – oneguy Aug 12 '17 at 22:35
• @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?. – dxiv Aug 12 '17 at 23:29
• oh, I'm sorry - I'll change it – oneguy Aug 13 '17 at 9:54

$2017^2=(2018-1)^2=2018^2-2\cdot2018+1$,
$2019^2=(2018+1)^2=2018^2+2\cdot2018+1$,
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