Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?

Question

There's a Brazilian 11 year old that allegedly developed an original method of finding natural square roots and is being marketed as some sort of genius for it. The method is even being called "the Julia Regression" by national media. It is sort of impressive for a human that age to notice this kind of relations, but the method per se seems rather dull to me:

N is given to be the square of some unknown natural number a, which we want to find.

Guess some natural number b to be this number.

Now compare b² with N

If (b² = N), then it's done

If (b² < N), you recursively sum b with its successor b+1 to b², that is:

1st try: b² + (b + (b+1))

2nd try: b² + (b + (b+1)) + ((b+1) + (b+2))

3rd try : b² + (b + (b+1)) + ((b+1) + (b+2)) + ((b+2) + (b+3))

Until you eventually find it

If (b² > N), you do the same but subtracting (b + (b - 1)) from b²

1st try: b² - (b + (b-1))

2nd try: b² - (b + (b-1)) - ((b-1) + (b-2))

3rd try : b² - (b + (b-1)) - ((b-1) + (b-2)) - ((b-2) - (b-3))

Until you eventually find it

Example: For the number 256, I'll guess 13. The square is 169. Now I'll sum: 169+13+14 = 196. Still not the result.

196+14+15 = 225

225+15+16 = 256

That's it, you get it

Explanation: this comes from the trivial observation of the manipulation of the product:

A x A = A²

A x (A+1) = A² + A

(A+1) x (A+1) = A² + (A) + (A +1) = (A+1)²

And eventually you run through all the integers. Same with subtraction. What do you guys think? Source from the main newspaper of the country:

https://g1.globo.com/jornal-nacional/noticia/2023/11/10/aluna-de-11-anos-ajuda-a-desenvolver-formula-para-descobrir-raiz-quadrada-de-uma-nova-maneira.ghtml

Answer 1

This is not a new method. It actually comes from:

$$(a-b)(a+b)=a^2-b^2$$

For consecutive perfect squares, it is $a-b=1$.

For instance,

$$196 - 169 = (14-13)(14+13)$$ $$196 - 169 = (14+13)$$ $$196 = 169+14+13$$

In general:

$$a^2 = b^2+a+b$$

We know that $a=b+1$: $$a^2 = b^2+(b+1)+b$$

But if an 11 year old kid noticed this she must be pretty talented.

Why do I say it is not a new discovery? Because see Fermat Factorization method that uses same property for factoring integers.

R.I.P Fermat (1665)

Answer 2

This is not a new method. It's essentially:

1. guess
2. if the guess was too small, guess one bigger (or vice versa)
3. repeat

This is just a bit better than blind trial-and-error, and would have been obvious to anyone computing square roots thousands of years ago. For example, a much better technique of the same basic sort is the so-called Babylonian method, which has been recorded as far back as CE 60.

Let's use the Babylonian method for your example. We are given $256$, we guess $13$, and find $13^2 = 169$. This is too low. Instead of trying $14$, the Babylonian method tells us that our guess was too low by about $$\frac{256-169}{2\cdot 13} = \frac{87}{26}$$

which is around $3$. So our next guess should be $13+3= 16$ and this hits the answer on the second try instead of the fourth.

Answer 3

It looks like a geometrical insight: the total surface equals the brown, plus two times one yellow plus the green: (the yellow ones are $1$ large).

The fact that a child (11 years old) has seen this, is an achievement indeed: it shows an insight which most children lack at that age, certainly if (s)he sees and iteration which might lead to an interesting finding.
... and that's it!!!

In my youth, I have calculated the total number of angles in a polygon ($(180 - \frac{360}{n}) \times n$), I have invented the Pascal triangle two years before I learnt about it in school (just by calculating the powers of $11$), at age $14$ I had serious insight in $a^p \mod p$, which has almost led me to the small theorem of Fermat, and I studied mathematics indeed, but so what?
I failed civil engineering, I failed a year at university, even while learning mathematics, I am now working for 25 years as a normal software engineer, ...

I just mean: that child might have insight in mathematics indeed, and this might lead to a college degree, but don't expect that child to become the next Einstein, Mozart or Newton for such a small achievement.

Answer 4

The proof for this method can be explained simply via telescopic cancellation method.

Consider a number $n^2=N$ whose square root $n$ we want to determine, then:

\begin{align*} n^2 - (n-1)^2 &= n + (n-1) \\ + \quad (n-1)^2 - (n-2)^2 &= (n-1) + (n-2) \\ +\quad (n-2)^2 - (n-3)^2 &= (n-2) + (n-3) \\ \quad ... \\ + \quad (b+2)^2 - (b+1)^2 &= (b+2) + (b+1) \\ + \quad (b+1)^2 - b^2 &= (b+1) + b \\ \hline n^2 &= b^2 + (b + (b+1)) + ((b+1) + (b+2))+...+ ((n-1) + n) \end{align*} which is the same as your answer for $b^2<N$.

For $b^2>N$, do the same approach but reverse.

So, it is not a new method since this trick has been known to people for thousands of years now.

Answer 5

Another way to understand the process, and many of the already posted answers, is the observation that a square number is the sum of a series of consecutive odd numbers beginning with $1$.

If your initial guess is too small (or too large), adding (or subtracting) sequential odd numbers will afford you the next larger (or smaller) square, and you will step your way to the correct result. For the reasons others have posted, the correct odd numbers to add (or subtract) begin with $b+(b+1)$ (or for subtraction, $(b-1)+b$)

Answer 6

I don't know how novel it is, but as other answers have said, it's not very useful. It's feasible only for small numbers, and furthermore it only works if the square root is an integer. If you want to know how to find the square root of perfect squares, you can just memorize the perfect squares. You could also memorize the following table:

c  f(c)
1   4
4   3
5   0
6   1
9   2

Given a number N, you can find the square root with the following steps:

1. Take N/100 rounded down, take the square root, round that down. Call that a. That is, a = int(sqrt(int(N/100)))

2. Concatenate a with 5. Call that number b. Square that number. Compare it to N.

2b. There's a trick to squaring numbers that end in five. Take a, multiply it by a+1. Then concatenate that with 25.

3. Take the last digit, apply the function to it. If b^2 > N, then add it to b. Otherwise, subtract it. That is, $\sqrt(N) = b\pm f(c)$.

So for 256, we divide by 100 and round down, and get 2. We take the square root and round down, and get 1. We append 5 to that and get 15. We square it and get 225, which is less than 256, so we know that when we get to step 3, we'll be adding rather than subtracting. The last digit is 6, f(6) = 1, so we add 1. This gets us 16.

There's an even easier way to get cube roots. Memorize these two numbers: 28, 37. Now follow these steps:

1. Divide by 1000, round down, take the cube root, round down. Call the result a.

2. Look at the last digit of the original number. If that digit appears in one of the numbers I asked you to memorize, take the other digit in that number. Otherwise, keep the original digit.

3. Take a and concatenate with the result of step 2.

For example, suppose you have 4913. Step 1 gets you 1. The digit 3 appears in 37, so we take the other digit, 7. 1 and 7 gets you 17. That's the cube root.

Answer 7

You got it wrong. She inspired her teacher to find this:

Suppose we want to find out that 21904 = 148²

We guess 99² and get 9801

We miss by |21904 - 9801| = 12103

Dividing by twice the guess we get 12103 / 99 / 2 ~= 61.13

The divisors of 12103 that fall nearnest to 61.13 are 49 and 91, since 12103 = 19 * 13 * 7 * 7

We'll try both in the formula (|diff| ± divisor²) / 2(divisor)

We'll use the sum version because we guessed lower.

(12103 + 49²) / 2(49) = 148 (correct)

(12103 + 91²) / 2(91) = 112 (incorrect)

This is what he did.

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Now, geometry raises two questions:

Why calculate (12103+49²)/2(49) instead of 99+49?

Why calculate a divisor above 61.13 if 2 * 49 * 99 + 49² > 2 * 49 * 99?