Between which two integers does $\sqrt{2017}$ fall?

$\sqrt{2017}\approx\sqrt{2000}=20\sqrt{5}\approx 20\cdot 2.236 \approx 45$ and $$44^2 = 1936,\qquad 45^2=2025$$ hence $\sqrt{2017}\in\color{red}{\left(44,45\right)}$.


$44^2 = 1936 < 2017 < 2025 = 45^2$.

Really, I don't think there's much to this one except for "try squaring small integers until you find the right ones".


You could use the root extraction algorithm to find it directly. It's sort of like long division.

  1. Starting from the decimal place, divide the number into pairs of digits. So $20\,17$.
  2. Find the largest integer whose square is less than the first pair. $4^2 < 20 < 5^2$. This is the first digit of the square root.
  3. Subtract off the square and bring down the next two digits. $20 - 4^2 = 4 \Longrightarrow 417$. This is our remainder.
  4. Find the largest $d$ such that the product $(20\cdot r + d)\cdot d$ is less than the current remainder, where $r$ is the part of the root already found. $(20\cdot 4 + 4)\cdot 4 < 417 < (20 \cdot 4 + 5)\cdot 5$, so $d= 4$.
  5. Add $d$ to the end of the digits already found, subtract the product from the current remainder, and bring down the next two digits. So we have 44 for our root and $417 - 84\cdot4 = 81 \Longrightarrow 8100$ for our new remainder.
  6. Repeat 4 and 5 until you have enough digits or the remainder and all remaining digits of the number are zero. Since we now have enough digits for the integer part, we can stop here.

So $44 < \sqrt{2017} < 45$.

I do want to comment on one suggestion you had, though.

However, 2016 (the number before it) and 2018 (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares

$2016 = 2^5\cdot3^2\cdot7$, which is divisible by a lot of perfect squares. Trial division by some of those perfect squares may result in a quotient that is itself close to a perfect square, which would give a guess as to what $\sqrt{2016}$ is. $2016 = 2^2\cdot504 = 3^3\cdot224 = 4^2\cdot126 = 6^2\cdot56 = 12^2\cdot14$. We should then recognize $224 \approx 225 = 15^2$ and $126\approx 121 = 11^2$. This gives $44^2 < 2016 < 45^2$, so clearly $44^2 < 2017 < 45^2$ as well.