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

Question

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

Since $2017$ is a prime, there's not much I can do with it. 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, so if I multiply them by a number to make them perfect squares, they're no longer close to $2017.$ How can I solve this problem?

Update: Okay, since $40^2 = 1600$ and $50^2 = 2500$, I just tried $45$ and $44$ and they happened to be the answer - but I want to be more mathematical than that...

Answer 1

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

Answer 2

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

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

Answer 3

it is $$44<\sqrt{2017}<45$$ since $$44^2=1936$$ and $$45^2=2025$$

Answer 4

As to the next question, Pell -1 and Pell +1..Hmm, seems to be showing the exponents incorrectly.

$$ {106515299132603184503844444}^2 - 2017 \cdot 2371696115380807559791481^2 = -1 $$

$$ 22691017898615873418283839489716246568157231499338273^2 - 2017 \cdot 505243842362839347335084683756885179819279000763128^2 = 1 $$

So, $\sqrt {2017}$ is above $$ \frac{106515299132603184503844444}{2371696115380807559791481} $$ and below $$ \frac{22691017898615873418283839489716246568157231499338273}{505243842362839347335084683756885179819279000763128} $$

parisize = 4000000, primelimit = 500509
? a = 106515299132603184503844444
%1 = 106515299132603184503844444
? b = 2371696115380807559791481
%2 = 2371696115380807559791481
? a^2 - 2017 * b^2
%3 = -1
? 
? c = 22691017898615873418283839489716246568157231499338273
%4 = 22691017898615873418283839489716246568157231499338273
? d = 505243842362839347335084683756885179819279000763128
%5 = 505243842362839347335084683756885179819279000763128
? c^2 - 2017 * d^2
%6 = 1
? 
? 
? a^2 + 2017 * b^2 - c
%7 = 0
? 2 * a * b - d
%8 = 0
? 
? 

Answer 5

Knowing the answer lies between $40$ and $50$ the rational thing to do would be to try the number in the middle of $40$ and $50$ i.e $45$.

If you don't feel comfortable with multiplying $45\cdot 45$ you can use the formula for $(40+5)^2=40^2+2\cdot 40\cdot 5+5^2=1600+400+25=2025$

Now you can use that $45^2-44^2=(45-44)(45+44)=89$ so one can see that $45^2>45^2-7=2017>45^2-89$.

It helps speed up a little of calculation for contests and such.

Answer 6

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.

Answer 7

Try graphing $\sqrt x$ for $x \geq 0$. You should see a fairly smooth curve that goes upwards, which means that if $a < b$ and they're both positive, then $\sqrt a < \sqrt b$.

From this, it's clear that the integers you want are $\lfloor \sqrt{2017} \rfloor$ and $\lceil \sqrt{2017} \rceil$. A calculator readily tells us that $\sqrt{2017}$ is approximately 44.911, so the answer is 44 and 45.

If you really want to do it by prime factorization, look at the divisors of 2016. Notice that $2016 = 42 \times 48$. Then $43 \times 47 = 2021$ and $44 \times 46 = 2024$, which should strongly suggest 45 is the greater of the integers you're looking for.

Answer 8

I can only imagine this was intended to be about $$ (10a + 5)^2 = 100 a (a+1) + 25, $$ $$ 15^2 = 225, $$ $$ 25^2 = 625, $$ $$ 35^2 = 1225, $$ $$ 45^2 = 2025. $$ Then $$ 44^2 = 2025 - 2 \cdot 45 + 1 = 2025 - 90 + 1 < 2017. $$

EXAMPLE: factor $10001 = 10^4 + 1$

$$ 105^2 = 11025 $$ $$ 105^2 - 10001 = 1024 $$ $$ 1024 = 32^2 $$ $$ 10001 = 105^2 - 32^2$$ $$ 10001 = (105 - 32)(105 + 32) = 73 \cdot 137 $$

Answer 9

For a rough estimate, I'd first divide by $100$ and think about $20.17$, for $\sqrt{2017} = 10 \sqrt{20.17}$. In fact, I would just consider $\sqrt{20}$:

You know $4^2=16$ and $5^2 =25$, so $\sqrt{20}$ is between $4$ and $5$, and is in fact close to the middle of them, i.e close to $4.5$. Hence, $\sqrt{2017}$ will be close to $45$.

So, see what $45^2$ is ... and proceed from there.

Answer 10

You can use the Newton square root method wikipedia for integers:

$$x_{n+1} = \frac 1 2 \left(x_n+ \frac S {x_n}\right)$$

Let us start with a crappy guess

1. $x_1 = 500$:
2. $x_2 = \frac{1}{2} (500 + 2017/500) = 252$
3. $x_3 = \frac 1 2 (252 + 2017/252) = 130$
4. $x_4 = \frac 1 2 (130 + 2017/130) = 72.7$
5. $x_5 = \frac 1 2 (73 + 2017/73) = 50$
6. $x_6 = \frac 1 2 (50 + 2017/50) = 45$

Now as the iterations seem to converge we can try $45^2 = 2025$ and we have our answer.

Answer 11

You can also use the least significant digits to get your bearings. Since $2016 \equiv 16 \pmod{100}$, if $2016$ is a perfect square, then $n$ in $n^2 = 2016$ is an integer satisfying $n \equiv 4, 6 \pmod{10}$. Clearly $n = 4$ or $6$ is too small.

Then, working our way up, we get $(196, 256), (576, 676), (1156, 1296), (1936, 2116)$, the last two corresponding to $44$ and $46$. Of course $45^2 \neq 2017$, but maybe it's $2025$. Notice then that $2116 - 2025 = 91$ and $91$ is the $45$th odd number. Likewise, $2025 - 1936 = 89 = 2 \times 44 + 1$, so it checks out.

So the answer is $44 < \sqrt{2017} < 45$.

Answer 12

There are well enough good answers here, but no one's suggested this method yet, so I'll add it.

Method $1$ - (Secant Approximation)

We can take two simple perfect squares that straddle $2017$. We're just getting a rough estimate, so we'd prefer numbers to be easy to work with than near $2017$. For example, $1600\leq 2017\leq 3600$. The secant line passing $(1600,40)$ and $(3600,60)$ should approximate $\sqrt{x}$ in the interval $[1600,3600]$. This gives us: $\sqrt{x}=24+\frac{x}{100}+\varepsilon_1(x)$. * So, $\sqrt{2017}\approx44$. Finally, we can check the squares of $43,44,45,\ldots$ and find our answer.

Method $2$ - (Mean of $2$nd Degree Taylor Polynomials)

We can make method $1$ slightly more accurate, though it's not really vital to do so. The $2$nd degree Taylor expansion of $\sqrt{x}$ at $x=a$ is $T_a(x)=\sqrt{a}+\frac{x-a}{80}+\frac{(x-a)^2}{8a\sqrt{a}}$. Then the mean of $T_{1600}$ and $T_{3600}(x)$ should be a good estimate for $\sqrt{x}$ near the centre of the interval $[1600,3600]$. Then $\sqrt{x}=60+\frac{x-2600}{80}-\frac{1}{2\cdot8}\left(\frac{(x-1600)^2}{40^3}+\frac{(x-3600)^2}{60^3}\right)+\varepsilon_2(x)$.** Hence, $\sqrt{2017}\approx45$ and we can check the squares of nearby integers, as in method $1$.

These methods would also work for any $2000<x<3000$ and could easily be adapted for other values of $x$. They also give a convenient way of finding an initial guess, for methods such as Newton-Raphson (detailed in @mathreadler's answer).

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Accuracy of Methods

* The error term reaches its maximum at $\varepsilon_1(2500)= 1$. So, in the worst case scenario, we'd need to check the $3$ numbers $(y-1),y,(y+1)$, where $y$ is the estimate of $\sqrt{x}$ from method $1$ and where $x\in[1600,3600]$.

** For $x\in(1769,3110)$, we have $\varepsilon_2(x)<0.67<\varepsilon_1(x)$ but for $x<1769$ or $x>3110$, we have $\varepsilon_2(x)>\varepsilon_1(x)$. In other words, method $1$ is more accurate than method $2$ in the centre of $[1600,3600]$, but the opposite is true near the bounds of the interval. However, since we're interested in the centre of the interval, this is good. The error term, $\varepsilon_2(x)$, reaches a minimum of $0$ around $x=2351$.