Going back a little, one part of elementary number theory that’s intimately connected to Dirichlet’s theorem on units is the Pell equation, *a*^2 – *nb*^2 = 1. Normally, *n* is restricted to be positive and not divisible by any square greater than 1; if *n* < 0 then the equation is trivial, and if *n* = *mc*^2 then the equation just becomes *a*^2 – *m*(*bc*)^2 = 1.

By Dirichlet’s theorem, the unit group of a real quadratic field **Q**(SQRT(*n*)) is **Z** * W, where W is the group of roots of unity. The only real roots of unity are 1 and -1, so W is just {1, -1} = C(2) = **Z**/2**Z**, and it’s fairly easy to show that the group of all positive units is **Z**. That already gives solutions to the Pell equation other than the trivial one *a* = (+/-)1, *b* = 0 – but note that actually finding a solution is more difficult.

Given that, it’s interesting to ask what the equivalent of 1 in **Z** is, i.e. what the smallest unit in **Q**(SQRT(*n*)) that’s greater than 1 is. If we’re already given a unit *a* + *b*SQRT(*n*), we can always calculate the smallest unit as a root of *a* + *b*SQRT(*n*). If the smallest unit is *c* + *d*SQRT(*n*), then expanding (*c* + *d*SQRT(*n*))^*m* shows that all terms containing SQRT(*n*) also contain *d*, so that *d* divides *b*. Also, if *m* is odd then all rational terms contain *c* so that *c* divides *a*, while if *m* is even then *c* divides *b*. This gives finitely many possibilities for *c* and *d*, which can be checked directly (caveat: if *n* = 1 mod 4, then this needs to be slightly tweaked to account for the fact that *a* and *b* can be half-integers).

I’ll spare you the derivation of non-trivial units – if you’re interested, just reread the post proving Dirichlet’s theorem, play with *b* and *h* a little, and thank the FSM you only have to deal with one linearly independent unit. But from the above, it immediately follows that the smallest unit in **Z**[SQRT(2)] is 1 + SQRT(2). For proof that the smallest unit in **Z**[SQRT(3)] is 2 + SQRT(3), note that *d* = 1 or else *c* + *d*SQRT(3) <= 2 – SQRT(3) < 1, and then *c* = 2, because N(1 + SQRT(3)) = -2 != (+/-)1.

A few eons ago, when reflecting on the problem with proving that there are no integer solutions to *x*^2 – 2 = *y*^3, I said that every unit of **Z**[SQRT(2)] is (+/-)1 times a product of some copies of 1 + SQRT(2) or its inverse, -1 + SQRT(2). This is the proof of that. In fact, once you note that 1 + SQRT(2) is a unit, you don’t need Dirichlet’s theorem; any unit not of the above form will induce a unit between 1 and 1 + SQRT(2), which the argument with *c* and *d* is enough to show is a contradiction.

Although a unit *a* + *b*SQRT(*n*) can have norm -1, whereas the Pell equation requires the norm to be 1, we can always use (*a* + *b*SQRT(*n*))^2, which is a unit of norm (-1)^2 = 1. But this suggests looking at a related equation, *a*^2 – *nb*^2 = -1. This equation isn’t always solvable – for example, if *n* = 3, then the smallest unit has norm 1 and so does the unit -1, which implies that so do all units.

In fact, if *n* = 3 mod 4, then the equation is never solvable, because mod 4 is becomes *a*^2 + *b*^2 = 3, which is impossible. This generalizes to the case where *n* is even divisible by a prime *p* that’s equivalent to 3 mod 4, because mod *p* the equation is *a*^2 = -1, which is again impossible (proof: if *a*^2 = -1 mod *p* then *a* is a primitive 4th root of unity mod *p*, which is impossible because the multiplicative group mod *p* is C(*p*-1), whose size is not divisible by 4).

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