Finding unique number properties

Question

To create my puzzles, I often use the numerical properties of the integers. However, as of recently, I feel like I am running out of properties to use.
So, why not make it a sort of game to find interesting properties for integers.

What are valid properties?

1. They are a formula, which is only allowed to use integers, symbols, and the operators $\{+,-,\times,\text{^}\}$. You may either use one equality signs or up to 2 inequality signs ($<$ only, however)
2. the property must be unique, that only a finite set of integers can satisfy it.

Restrictions imposed on the numbers:

• You may only use integers. 0, and all whole negative numbers are allowed. However, fractions and irrational numbers such as $\sqrt 2$ are strictly forbidden. This also applies to the properties.
• numbers of which you want to display their properties are represented by symbols, preferably $a-z$
• Within the same property, any two symbols may NOT be equal to each-other or to any integer present in the equation.

Since I want to aim towards certain aspects of these properties, there will also be a scoring system involved:

• The equality/inequality sign rewards you 1 point. If you use 2 inequality signs, you get 2 points.
• Each operator you use costs you 1 point. A pair of parenthesis costs 2 points.
• Each number (NOT symbol) also costs you 1 point
• Each NEW integer reduces your score by its value. For example, if you are using the number 3 twice, you will lose 3 points for using 3 in general, and 2 more points for using 2 numbers.
• The rewarding system is a bit more complex for the symbols:
  • let's say that a symbol only has 1 possible value that can fit in it. Therefore, it will reward you as many points as the absolute value of that number. This means that negative numbers are good! (this is the base step)
  • let's say that there are multiple symbols that each have only 1 possible value that can fit. Therefore, you are rewarded the sum of their absolute values. (this is the AND step)
  • lastly, if there are multiple options for those symbols, you are rewarded the average of the situations. (this is the OR step)
  • a generic method of calculation the reward if to do things using the order of operations: first determine the bases, then the AND steps, and lastly, the OR step.

Now for two of my favorite properties as examples: $$a \times a = a$$ This property has 1 equality sign, 1 operator, and 1 symbol. This means we are immediately rewarded 1 point but then penalized that same point by the operator. We have an initial score of 1. The solution for $a$ can be either 1 or 0, so it gives us the average of those two numbers, or 1/2. Therefore, the score of this formula is $0.5$

$$a^b=b^a$$ Quite the sneaky property. It has 1 equality sign and 2 operators, leaving us with an intial score of -1. yikes. However, there are 4 solutions for the equation: $\{(2,4),(4,2),(-2,-4),(-4,-2)\}$ After applying the base step, we get $[(2,4),(4,2),(2,4),(4,2)]$. After the AND step (which sums the first set of parentheses, we get $[6,6,6,6]$ which averages out to 6 by the OR step. Thesefore, the score of this property turns out to be 6-1=$5$, which is not bad.

How well can you do?

NOTE: I will also point out that the set 0 score is for the simplest conceivable property: $a+a=a$ which has the score of 0. Also, even though negative scores are possible, I am fine with them (unless they are ridiculously negative)

Answer 1

The following is not a mathematically interesting property, but it scores high.

$a^2 < a \times 2^{2^{2^{2^2}}}$

It uses:
$1$ (in)equality
$6$ operators
$2$ is the only the number value used
$6$ numbers
which gives a penalty of $1-6-2-6 = -13$.

The penalty is insignificant, because its solutions are the integers

from $1$ (inclusive) to $2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}$ (exclusive). These have an average of $2^{65536}/2=2^{65535} \approx 10^{19728}$.

This can extended in an obvious way - adding one $2$ and one operator which only gives two extra penalty points - for an even more outrageous score.

Of course I could instead simply use this property:

$a=2^{2^{2^{2^2}}}$

but that would be even less interesting.

Answer 2

I will also go ahead and place two other properties I know: $$a^a=a$$ It has 2 solutions: -1,1 (0^0 is considered NaN). The proof for this involves using the known function $^n\sqrt n$ and its properties. the initial score is 0, and then we add the symbol score of 1 (as (|1|+|-1|)/2=1) to get the total score of 1. $$a^2+2=b^{2+1}$$ There is one solution to this: $a=5,b=3$.
The Proof is here: www.normalesup.org/~baglio/maths/26number.pdf
Taking a look at the equation we see: 1 equality, 4 operators, the numbers 2,1 used, and 4 numbers total. Equality: +1, Operators: -4 (total -3), New numbers: -3 (total -6), total numbers: -4 (total -10), Symbols: 5+3 = +8 (total -2) if someone could prove that this other form of the equation works: $a^2+2=b^b$, that would be nice.