By sixth grade (I think), you’ve learned some basic facts about addition, multiplication, and exponentiation over the natural numbers: you’ve learned that addition and multiplication are commutative and associative, that multiplication distributes over addition, that 0 is an identity for addition and 1 is an identity for multiplication, and the following simplification rules for exponents:

1^{x} = 1

*x*^{1}* = x*

*x*^{y + z} =* x*^{y} ⋅ *x*^{z}

(*x* ⋅ *y*)^{z} = *x*^{z} ⋅ *y*^{z}

(*x*^{y})^{z} = *x*^{y ⋅ z}

Tarski called these the High School Identities (but I think we learn them before high school) and asked if every identity true in the natural numbers that just used addition, multiplication, and exponentiation followed from them.

The answer turned out to be “no.”

In 1981, the model theorist Alex Wilkie discovered that the identity

((1 + *x*)^{y} + (1 + *x* + *x*^{2})^{y})^{x} ⋅ ((1 + *x*^{3})^{x} + (1 + *x*^{2} + *x*^{4})^{x})^{y} = ((1 + *x*)^{x} + (1 + *x* + *x*^{2})^{x})^{y} ⋅ ((1 + *x*^{3})^{y} + (1 + *x*^{2} + *x*^{4})^{y})^{x}

is true for the natural numbers, but does not follow from the High School Identities.

But don’t be concerned that you have to memorize this new identity. It follows from the fact that 1 + *x*^{3} = (1 – *x* + *x*^{2}) ⋅ (1 + *x*) and (1 + *x*^{2} + *x*^{4}) = (1 – *x* + *x*^{2}) ⋅ (1 + *x* + *x*^{2}) and if we could express (1 – *x* + *x*^{2}) in our language, we would be able to prove Wilkie’s identity from the High School Identities. The only problem is the –*x*, since we don’t have additive inverses.

It’s tricky to just put in additive inverses as well, since that will necessitate adding the rational numbers, and then complex numbers, and then exponentiation is no longer single-valued. But Wilkie also proved that if we allow ourselves to factor expressions into polynomials with integer coefficients so long as those polynomials are always positive on the positive reals, then the High School Identities prove every true exponential identity.

For more information, see Wilkie’s article or a survey article by Stanley Burris and Karen Yeats.

$1+x^3=(1-x+x^2) \cdot (1+x$ — is that the same as the trick Cardano used to get cubics into reduced form? (that shows up in the story of $\sqrt{-1}$)

I’m missing something. Of course $3$ needs its $-3$—but how does merely requiring additive inverse necessitate $\sqrt{-1}$?