The other day someone asked:
- Is the product of 4 consecutive
positive integers always one
less than a square?
Good question. The answer is yes, and I solved it using a technique
the interlocutor didn't know:
OK. If you're interested in the actual algorithm then simply skip
to the end. However, I'm going to derive the algorithm, starting
with long division.
So let's start with the division problem, computing 999999 divided by 7.
Modern teaching doesn't use long division any more, apparently. Instead
they use a thing called "chunking". In fact, "chunking" is nothing more
that long division done with more guesses and less formality.
1
+-------------
7 | 9 9 9 9 9 9
7
---
2 9 9 9 9 9
|
|
So we start by ignoring everything except the most significant part. We
ask how many 7s we can take out from the first digit. The answer is just
one. So we have
999999 = 700000 + 299999
We can see that laid out formally at right.
That table then says that
9999999 / 7 = 100000 + 299999/7
We now iterated. We can remove 4 chunks of 7 from 29, so that's removing
40000 lots of 7 from 299999. Again, we lay that out in the tabular fashion:
1 4
+-------------
7 | 9 9 9 9 9 9
7
---
2 9 9 9 9 9
2 8
-----
1 9 9 9 9
|
|
So we are, at each stage, deciding what the next digit should be, multiply
by 7, subtract from the remaining amount, and then repeat. This is laid out
in a formal, tabular fashion. At each stage we have an intermediate sum
that makes sense.
1 4 2 8 . .
+-------------
7 | 9 9 9 9 9 9
7
---
2 9 9 9 9 9
2 8
-----
1 9 9 9 9
1 4
-----
5 9 9 9
5 6
-----
3 9 9
|
|
Here we can see that 999999 = 7 * 142800 + 399. At each stage we reduce
the remainder, subtracting a chunk of 7s from it.
This can be used for polynomials too. So, for example:
x^2 - 2x + 2
+---------------------------------
x^2 + x - 3 | x^4 - x^3 - 3x^2 + 8x - 5
x^4 + x^3 - 3x^2
------------------
-2x^3 + 8x
-2x^3 - 2x^2 + 6x
-------------------
+ 2x^2 + 2x - 5
2x^2 + 2x - 6
-------------------
1
|
|
As you can see, we can divide $x^4-x^3-3x^2+8x-5$ by $x^2+x+3$ and we
get $x^2+2x+2$ with a remainder of 1.
So we can perform long division with polynomials in exactly the same way as
we can perform long division with numbers, and in effect, it's just a formalised
way of doing "chunking."
So, on to Square Roots By Long Division ...
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