「新デノミ論」or, A New Currency Unit for Japan

Okay, so this is only marginally mathematical, but here's a link to my brilliant (?) proposal for resetting Japan's currency roughly at parity with the US dollar (as it was, not that you asked, back when the yen was adopted in 1871, a hundred years before I got here). The paper at the link is in Japanese; in it, by way of summary, I propose keeping the yen but introducing the "rin" (輪) as the new primary currency unit, with 1 rin equal to 100 yen (and thus, at the current exchange rate, roughly equal to US$1).

Link: 新デノミ論 (Shin denomi ron)

The product of the sums of two squares is…

This may not come as a great surprise to anyone else, but I was mildly pleased to stumble on the fact (?) that if two integers are both the sums of two squares, so is their product:

If

x = a² + b²

and 

y = c² + d²

then 

xy =

(a² + b²)(c² +d²) =

a²(c² + d²) + b²(c² +d²) =

a²c² + a²d² + b²c² + b²d² =

(a²c² + b²d²⁾ + (a²d² + b²c²⁾ =

(a²c² +²abcd + b²d²) + (a²d² -²abcd + b²c²) =

(ac + bd)² + (ad - bc)²

and by a similar procedure we can show that
xy = (ac - bd)² + (ad + bc)²

Thus xy is actually the sum of two pairs of squares, unless ad = bc or ac = bd.

Or so it seems to me.

Real square roots of –1

Real Square Roots of –1

Yes, well I quite like imaginary and complex numbers, starting with i and –i, the square roots of –1.
But did you know that –1 also has square roots that are real numbers?
Well it does.
Sometimes.
In modular arithmetic.

In particular, if p is prime and congruent to 1 modulo 4, then –1 has a pair of real integer square roots modulo p.

Why? 

Consider that there are p – 1 distinct non-zero numbers in modulus p. 

If p is prime, each has a unique reciprocal (allow me to spare you the proof, which is simple but I’d probably mess up). 

And the reciprocal of the inverse of each is the inverse of the reciprocal. 

So we have sets of four numbers, a, –a, b, –b, such that 1/a = b and 1/(–a) = –b. 

But if a = 1, b =1, and of course –a = –1 and b = –1. 

Now, if p ≡ 1 (mod 4), p – 1 is divisible by 4, meaning that 1 and –1 must be part of a foursome with another pair of numbers. 

And since these numbers cannot be their own reciprocals (only 1 and –1 can manage that), they must be reciprocal to each other. 

IOW, a = 1/(–a) and –a = 1/a. 

Whence a²= ­–1 and (–a)² = –1.

So there you have it. 

Playing with sums of two squares

Yes, well . . .

I recently happened to notice that if
(a^2) + (b^2) = 2p,where a, b, and p are all integers, then
((a+b)/2)^2 + ((a-b)/2)^2 = p.

For example, if we know that
5^2 + 1^2 = 26 = 2*13,
it follows that
((5+1)/2)^ + (5-1)/2)^2 = 13,
that is,
3^2 + 2^2 = 13.

Suppose, though, that
(a^2) + (b^2) = 3p
or, more generally,
(a^2) + (b^2) = op,
where o is an odd integer.

Is there some way of deriving from this a formula such that
(c^2) + (d^2) = p,
where c and d are also integers, with values determined by a and b?

Lots of powers

For every pair of complex numbers a and b, provided a is not 1 and neither a nor b is 0, there exists a complex number c such that a^c = b. In other words, there is a number c such that b is the cth power of a. Another way of putting this is that there is a number c such that a is the cth root of b. 

Given a and b, we can find c simply by calculating (ln b)/(ln a). Of course (?) the natural log function ln z is multivalued, so there will actually be an infinite number of solutions. If we want just one, we can use the formula c = (Ln b)/(Ln a), where Ln z is the principal value of the complex logarithm.

Division ±mod 11

OK, I've waited too long. Here's an example of modular division with a prime modulus, i.e., 11. By way of reference I also include the multiplication table. I have replaced 6, 7, 8, 9, and 10 with their negative values: –5, –4, –3, –2, and –1, respectively.

I won't attempt to explain this or comment on it, other than saying I think it's beautiful.

Modular division

Sorry, I've been very lazy about adding content here. In this entry I will consider modular division.

If the modulus m is not prime, some divisions have multiple results and others have none. Here, for example, is the division table for mod 10):

We see that 1, for example, is divisible only by itself, 3, 7, and 9. By contrast, 2 is divisible by every number except 5 and 0, but some of the results overlap: 2/2 equals both 1 and 6; conversely both 2/2 and 2/7 equal 6. The number 5 is rather special: It can be divided by every odd number; 5/5 can equal any odd number, and 0/5 can equal any even number; also, 5 shows up as a second quotient when 0 is divided by an even number.

In terms of regular, non-modular arithmetic, we may interpret the results as follows: 1/1=1 (the result shown in column 1, row /1) means that if an integer ending in 1 is evenly divisible by another integer ending in 1, the quotient will also end in 1, for example, 1/1, 11/1, 11/11, 111/1, and 121/11. And 6/8=2, 7 (column 6, row /8) means that means that if an integer ending in 6 is evenly divisible by an integer ending in 8, the quotient may end in either 2, as with 16/8, 36/18, and 96/8, or 7, as with 56/8, 126/18, and 136/8.

Where/how did I find these results? Well, the way I actually did it was slightly (?) messy, but the simple answer is that they can all be derived from the mod 10 multiplication table:

For example, consider the /2 row of the division table. To fill in the values, we must find 1/2, 2/2, 3/2, etc. In other words, we are looking for the values that satisfy a=1/2, b=2/2, c=3/2, d=4/2 and so forth. We can rewrite the equalities as 2a=1, b=2, 2c=3, 2d=4. etc. Having done so, we can look for a simply by running our eye down column 2 and finding the row(s) in which the result is 1, meaning that 2a=1. There aren't any. No number can be multiplied by 2 to give a result of 1 in mod 10.

With d=4/2, by contrast, if we follow the same procedure, rewriting the equality as 2d=4, and looking down column 2, we find a result of 4 in row 2 and again in row 7.

Comparing the two tables, we see that division by 1 (row /1 in the division table) is equivalent to multiplication by 1. This of course is not very surprising or interesting. Rather more interesting is that division by 3 turns out to be equivalent to multiplication by 7, and vice versa. Also, division by 9 is equivalent to multiplication by 9.

Now let's try switching to ±mod 10, meaning a notation using both positive and negative numbers (as introduced in my previous entry). This means replacing 9 with –1, 8 with –2, 7 with –3, and 6 with –4. For balance, let's add + signs to 1, 2, 3, and 4, and so as to leave no number unsigned, let's add ± signs to 5 and 0. Here's how the division table looks with this notation:

Just as with multiplication (see my previous entry), using ±mod notation gives all sorts of additional symmetry to the table. Above I wrote, "Division by 3 turns out to be equivalent to multiplication by 7, and vice versa. Also, division by 9 is equivalent to multiplication by 9." These results become less surprising here. In mod 10, 3*7 = 1, which we take as just one of the many results of multiplication. But in ±mod 10, we can easily derive (+3)*(–3) = –9 = +1. This strikes me as more satisfying, though of course it is simply a matter of using different notation. And the second result, concerning division and multiplication by 9, becomes downright trivial when presented as "division by –1 is equivalent to multiplication by –1."

So that's mod 10 division. Things get much more interesting, IMO, when we consider prime moduli, which I plan to do in my next entry.