This was a question which intrigued me when I was a student: is there a meaningful mathematical statement about finite fields which is proven by induction on the characteristic of a field?

Finally and many years later I got a partial answer: a meaningful statement about prime numbers proven by induction on the size of prime number in question.

This is a note *A Lemma on Divisibility* by Peter Walker in a recent issue of The American Mathematical Monthly 115 no. 4 (2008 ) p. 338:

(*) For all primes , implies or .

His proof uses division with remainder but not Euclidean algorithm for finding the greatest common divisor of two integers.

We say that a prime is *genuine* if it satisfies (*) for all integers and . We shall prove by induction on that all primes are genuine.

**Basis of induction. ** is a genuine prime number. Indeed if a product of two integers is even then at least one of the multiplicands is even.

**Inductive step.** Let be the least non-genuine prime number so that but does not divide and does not divide .

Using division with remainder, we can write and where . Of course, . If either or then divides or as required. If not, then both and are at least and can be factorised into primes: and (*existence* of factorisation can be proved earlier). Now and for some integer we have wher all , are less than and are therefore *genuine prime numbers*. Since none of , can divide , they divide and can be cancelled out one by one from the equation, leaving an obviously contradictory equality .

How do you know this fact: (*) if ab is even then one of a and b are even.

Of course this follows from unique factorization but the proofs I know of unique factorization require the divisibility lemma. Can you give an independent proof of (*)?

By:

Charles Wellson April 20, 2008at 1:34 am

(*) if is even then one of and are even.

Sketch of a proof: if is odd and is odd then and . Then

is also odd.

By:

Alexandre Borovikon April 20, 2008at 5:22 am

How do you know that ab might not be both odd and even, without deriving that fact from unique factorization? Well, one answer is that if m and n are integers for which 2m+1 = 2n, then 2(m-n) = 1, a case of the “contradictory inequality vp = 1” that you mentioned. By the way, the impossibility of vp = 1 unless v = p = 1 (for positive integers) follows easily from the Peano axioms and the recursive definition of multiplication.

By:

Charles Wellson April 21, 2008at 1:59 pm

[…] and beautitiful proof of a self-evident statement, like a proof of lemma in my old post Induction over prime numbers? Possibly related posts: (automatically generated)What is deep mathematics?Arwenâ€™s FREE […]

By:

Elusive boundary of proof « Mathematics under the Microscopeon February 24, 2009at 3:05 pm

@Charles: You know it from induction “step”. Contrary to popular opinion, induction with total history (if all primes less than p are genuine, then p is genuine) _doesn’t_ need a basis, since it is just a special case of a step: all primes less than 2 _are_ surely genuine (since there are none), so 2 is genuine by induction step.

Only inductions with bounded history (eg, if P(k-1) and P(k), then P(k+1)) need a basis, since you can’t conclude P(0) by the step if P(-1) doesn’t hold or even make sense in the context.

@Alexandre: So the real proof is even shorter, in fact, and the whole even/odd distinction is a special case of what’s going on in the step. Really, in the step we invoke theorem about division with remainder, which obviously answers second Charles’s question as well (when p is 2, remainder can be 0 or 1, and can’t be both).

By:

Vekyon February 25, 2009at 11:10 am