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If no calculator is allowed, and we want to find the square root of a square number if it is large and analyzing to prime factors is hard, how can one proceed? For example, what to do if the number is a square of a large prime number like 5329?
What are methods to perform this and what are pedagogical reasons to prefer one over the other?
You can use (basically) Newton's method.
Take $x_0>0$, and define $$x_{k+1} = \frac{1}{2} (x_k + \frac{n}{x_k}).$$ The seqeunce converges to $\sqrt{n}$.
And if one starts with $x_0= n$ one has, as soon as $|x_{k+1} - x_k|< 1$, that $\lfloor x_{k+1} \rfloor = \lfloor\sqrt{n} \rfloor$.
But the above is not very feasible for computing by hand, and there is better variant, if you just want to do this for perfect squares or test if something is a perfect square, or more generally compute $\lfloor\sqrt{n} \rfloor$. (It seems this is the case for you.)
You can do $$x_{k+1} = \left \lfloor \frac{1}{2} (x_k + \lfloor \frac{n}{x_k} \rfloor) \right \rfloor$$ only doing arithemtic with integers (including euclidean division). This will converge to $\lfloor\sqrt{n} \rfloor$ in a finite number of iterations.
See Integer Square Root on Wikipedia for further references.
I think this could be an interesting method to teach, as one can provide a good geometric motivation for it.
There is an algorithm to calculate square roots with paper and pencil (like the long division algorithm). See How to calculate a square root without a calculator.
A fancy way, for those who know a little calculus, is to linearize $f(x)=\sqrt{x}$ at some point near the number you want to square root. This can only be done when you can think of a square rootable number that isn't too far away, so it's not going to work if you're trying to square root a super huge number, but if you can think of one, it's very doable by hand, and very quick. It also has the added benefit of giving you a rational approximation, if that's something you want.
For example, to compute $\sqrt{2}$, we first have to come up with some other number that is reasonably close to $2$ which we do know how to square root. The first thing that comes to mind is $9/4$ (though there are many other choices that are even more accurate, e.g. such as $49/25$). We linearize $\sqrt{x}$ at $9/4$: $$\left.\frac{d}{dx}\sqrt{x}\right|_{x=9/4}=\left.\frac{1}{2\sqrt{x}}\right|_{x=9/4}=\frac{1}{2\sqrt{9/4}}=\frac{1}{3}$$ $$\mathcal{L}_{9/4}(x)=f^\prime\left(\frac{1}{3}\right)\left(x-\frac{9}{4}\right)+f\left(\frac{9}{4}\right)=\frac{1}{3}\left(x-\frac{9}{4}\right)+\frac{3}{2}$$ Now, we plug $2$ in there. $$\mathcal{L}_{9/4}(2)=\frac{1}{3}\left(2-\frac{9}{4}\right)+\frac{3}{2}=\frac{17}{12}$$ From here, if you want decimals, you can simply use long division. Numerically, $17/12$ comes out to be $1.41667$, while the actual value of $\sqrt{2}$ is $1.41421\ldots$. That's pretty close, and likely close enough to do the job for many on the fly calculations.
Use log tables and the identity $x^\alpha = e^{\alpha \log x}$.
From an educational point of view, that gives evidence of the usefulness of logarithms (you can calculate any power of $x$ using just a log table and multiplication) and teaches about transforming between problems. It also allows all roots to be calculated by the same method.
An intuitive approach to find the square root of A is to draw a square with the number (A) in the middle as the area, and the sides labeled x and y. We know that x times y equal A, and x is the square root of A when x = y. Begin with a rough estimate of x and calculate y as A/x. Now replace x with the average of the original x and y, and repeat.
For example, let's find the square root of 200. As a rough guess, let's begin with x = 15. Solve for y = 200/15 = 13.333. The new value for x, call it x1 = (15 + 13.333)/2 = 14.167. Now the new value y1 = 200/14.167 = 14.117. The square root of 200 is between 14.167 and 14.117. The average is 14.142. If we use x2 = 14.142, then y2 = 200/14.142 = 14.142, verifying that we have the square root of 200 with five significant digits.
This approach has the advantage of showing what a square root is visually, which can give a problem more meaning. The process makes intuitive sense, and it does not require mathematics beyond division.
Here is the method I used to solve the original poster's problem in about three seconds:
Square root (5329):
Steps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude:
1a) Start with an estimated answer of 1.
1b) Find the order of magnitude of the answer. Make a copy of the argument (5329). Shift the decimal of the copy of the argument by two places at a time, to get a number between 1 and 100. Shift the decimal of the estimated answer by one place at a time, in the opposite direction. For example, 53.29 and 10.
1c) Assume that the student knows the squares of the integers between 0 and 10 by heart. What is the closest such square? For example, 49.
1d) Multiply the estimated answer by the square root of the closest square. For example, square root (49) = 7, so the new estimated answer is 70.
Step 2 checks the estimate:
2) Square the current estimated answer. For example, 70 * 70 = 4900.
Steps 3a-d refine the estimate using uses (a+b)^2 = a^2 + 2*a*b + b^2. This is a special case of Newton's method:
3a) Subtract the value found in step 2 from the original argument, to yield a remainder. For example, 5329 - 4900 = 429. Keep track of the sign.
3b) Make a copy of the estimated answer. Multiply the copy by 2, to yield a double estimate. For example, 70 * 2 = 140.
3c) Divide the remainder by the double estimate, to yield an adjustment. Round off as convenient. Keep track of the sign. For example, 429 / 140 ~ 3.
3d) Add the adjustment to the estimate (not the double estimate). We kept track of the sign, so that we could correctly choose whether to add or subtract. For example, 70 + 3 = 73. For subsequent calculations, this value replaces estimate.
Repeat steps 2 - 3 until the answer is close enough. For example, 4900 + 420 + 3*3 = 5329, so the square root of 5329 is 73.
Some advantages of this method:
I can do it in my head.
It only requires keeping track of a few numbers.
It quickly produces an estimated answer.
It is a special case of Newton's method.
Each time through the loop, the student can round off the adjustment to a convenient decimal or fraction.
Every iteration, the student checks their work.
Three worked examples:
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 5329 1
1b 10 53.29
1c 7*7=49
1d 70
2 5329 70 4900
3a 5329 70 4900 429
3b 5329 70 4900 429 140
3c 5329 70 4900 429 140 3
2 5329 73 5329 73*73=4900+420+9
Done!
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 739 1
1b 10 7.39
1c 3*3= 9
1d 30
2 739 30 900
3 739 30 900 -161 60 -3
2-3 739 27 729 10 54 0.2
2-3 739 27.2 739.84 -0.84 54.4 -0.015
2-3 739 27.185 739.024225 -0.024225 54.37 -0.0005
2-3 739 27.1845 738.99704025 54.369 0.00005
2 739 27.18455 738.9997587025
...Stop when satisfied...
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 4285 1
1b 10 42.85
1c 6*6=36
1d 60
2-3 4285 60 3600 685 120 5
2-3 4285 65 4225 60 130 0.4
2-3 4285 65.4 4277.16 7.84 130.8 0.06
2-3 4285 65.46 4285.0116 -0.0116 130.92 -0.0001
2-3 4285 65.4599 4284.99850801 130.9198 0.00001
2 4285 65.45991 4284.9998171081
...Stop when satisfied...
