What's asserted without proof can be denied without proof. Euclid of Megara (c.450-374 BC)
The interest I have to believe a thing is no proof that such a thing exists.
François-Marie Arouet, known as
Voltaire (1691-1778)

(2018-01-10) Proof by Inspection: The most elementary type of proofs.
This can be invoked only if there are finitely many cases to check.

An example is no proof. (Yiddish proverb)

In ordinary mathematical discourse, we may say that something is true by inspection
when there are only finitely many possible instances and the statement is easily checked for
every single one of them.

In the past, we'd only make the claim when there were only a few cases to check;
possibly too many to list, possibly a tedious task but not an overwhelming one.
In the computer era, we may also claim that something is true by inspection
when a (relatively simple) computer program has checked all the possible cases.

Arguably, proper mathematics consists in producing
proofs of statements applicable to infinitely many things, not statements true
by inspection.

Infinity marks the beginning of proper mathematics.

(2007-01-09) Only "negatives" are worth proving.
"You can't prove a negative" is a proverb about tests, not proofs.

To a mathematician, proofs are not restricted to mere tests.
Arguably, the aforementioned word negative doesn't mean much, since
grammatical form is only incidental.
To play along, we'll dub the first of the following sentences positive
and the second one negative :

A square can be the sum of two nonzero squares.

A cube cannot be the sum of two nonzero cubes.

Both statements are true in the realm of integers.
The first one can be proved by just one example
(the most popular of many
is 25 = 16+9).
On the other hand, the second statement tells that counterexamples do not exist...
The evidence for such an affirmation can only be a detailed piece of rigorous reasoning;
a proper proof.
A lack of solutions can't be established by many failed attempts.
That's probably what was meant by whoever coined the above proverb,
which doesn't apply to mathematical proofs.

Since 1 < Ö2 < 2,
if a positive integer n was making
nÖ2
an integer, the smaller positive integer
m = (Ö2-1) n
would make mÖ2 an integer also!

Another proof
invokes the concept of divisibility.
It may be easier and more intuitive, but it's less elementary
(it relies on more previous knowledge).

(2012-06-11) Proof by induction
(Gersonides, 1321)
The basic way to prove a statement about infinitely many things.

The elementary type of induction
(as taught at the high-school level) pertains
to integers: To establish that some statement P(n)
is true for all nonnegative integers, you only have to
show that:

P(0) is true.

Assuming the truth of P(i) for every i < n
(the so-called induction hypothesis)
it can be proved that P(n) is true.

Although the first part is pleonastic (it's only a special case
of the second part with n=0, with a vacuous induction hypothesis)
it's useful to keep it, since
the proof of the second part would otherwise almost always start
with a distinction between n=0 and the other cases.

That type of elementary induction could be reformulated to apply to the elements
of any countable set.

However, the general concept of induction
(sometimes known as structural induction )
has no such restrictions. Loosely stated:

If something is true of the simplest things and can be shown
to hold true of more complex things by assuming it's true of simpler
ones, then it holds true of the most complex things.

In this context, it's just assumed that those "most complex things"
are structually composed of simpler ones in a predefined way.
For example, Conway's surreal numbers
are simply built from simpler surreal numbers.
Structural induction can thus be used to establish the
validity of a statement about all surreal numbers
(there are uncountably many of them)
in a way that does not reduce to simple induction on integers.

(2007-01-16) Stochastic proofs are wrong with
vanishing probability.
Sometimes, an elusive truth is reinforced by many failures to attack it.

One celebrated example is the iterated
Rabin-Miller test
which tells (beyond the shadow of a doubt) whether a large number
is prime or not, without actually proving anything
when that number happens to be prime...
For a composite number,
each iteration stands a substantial chance (over 75%) of proving it's
not prime.
Thus, if several iterations fail to provide such a proof,
we may be very confident that the number is indeed prime
(the probability of error decreases
exponentially with the number of iterations).

Another example consists in determining
whether a (large) finite group
is cyclic (knowing the factorization into primes of its
order).
A finite group is cyclic if and only if it
has a primitive root.
It turns out that a random element of a cyclic group is
primitive with a fairly large probability
(and it can be proved to be primitive very efficently
if the prime factors of the group's order are known).
Thus, if many random elements turn out not to be primitive,
then the group is "almost surely" not cyclic.

(2007-01-16) Heuristic Arguments
Establishing the likelihood of a conjecture with an approximative proof.

For example, I argue (against the dominant opinion)
that there are probablyinfinitely many Wieferich primes,
although only two of them are known (in spite of
great efforts to find a third).

A proper heuristic argument is not a hasty generalization.
It's actually a strict mathematical proof about a modified
problem, where part of the original mathematical structure is
substituted with a probabilistic model.
Quantitative conclusions from such a model can be enlightening
while an exact solution to the original problem remains elusive.
This may be construed as "relaxing" some mathematical
constraints while retaining the problem's essential aspects.

A good heuristic argument must be supported with convincing
justifications
of the probabilistic assumptions underlaying the model.
A heuristical argument is never foolfproof (or else it would be a proper
mathematical proof) but it should be nearly so...
The qualifier "heuristic" shouldn't be an excuse for sloppiness !

The accepted heuristic arguments gave the
wrong answer for a paradoxical result which has now been proved
rigorously:
Maier's theorem (1985).

(2013-05-01) Computer-Assisted Proofs
Appel & Haken used a computer to prove the 4-color theorem (1976).

They were able to reduce the general case to 1936 special cases
that could not be reliably checked by hand.
The fact that a legitimate proof of a major theorem had not
been verified by a human being raised eyebrows at the time.

At a more modest level,
I once produced (2002-07-08)
a satisfying proof of a long-standing conjecture of mine by reducing it to
1172 very simple computations.
In that case, it would have taken only a few hours to do so by hand,
but I must confess that I never did.
There was no need for that,
as I could write a tiny computer program to do it for me.

The fact that I had previously enlisted the help of a computer to find
the magic modulus 1171 (and other larger suitable ones) is irrelevant
to the final proof. The creativity involved, if you must call it that,
resides entirely in the idea that such a modulus might exist.

Such proofs are getting more and more common these days.
Admittedly, they ultimately rely on the proper working of a computer.
However, the correctness of the computerized procedure
has to be established the old-fashioned way. Therein lies the crux of the proof.

(2009-06-21) Ruling out proofs
Facts which can't be established by some or all types of proofs.

Generations of mathematicians have attempted to prove Euclid's
fifth
postulate of plane geometry from the other four axioms of Euclidean
geometry.
It is now known that such a proof is not possible.
The reason why this is so is rather subtle:

If Euclidean geometry
(including the fifth postulate about parallel lines)
is at all consistent, then it can serve as a framework to
describe other surfaces besides a plane.
One such surface is the sphere...

The geometry of the surface of a sphere provides one example
where the first four axioms of Euclid are verified with
suitable redefinitions of the concepts involved
("points" are actually pairs of diametrically opposite locations
and "lines" are great circles). Yet, the
fifth postulate is not verified, as all "lines" intersect
(there is no such thing as two "parallel" great circles).

Therefore, the fifth postulate cannot
be a consequence of the other four.
Note that this conclusion can be reached without settling
the question of whether the Euclidean postulates
are consistent or not. We just note that if
they are consistent, then a consistent "model" can be
constructed (spherical geometry)
where the fifth one is false.
Hence, that fifth postulate is truly an independant
axiom which may be assumed to be true or false.

In examples of lesser historical significance,
similar arguments can be used to rule out some types
of proofs for a given statement.
For example, "Fermat's last theorem" can be shown to be false
within certain "models"
(involving beasts like p-adic integers).
This shows that it is a so-called global
statement whose proof must involve some peculiar
property of the rational integers besides
ordinary algebra, ordering
and divisibility by finitely many prime numbers.
The proof must involve something very specific to the integers,
like the validity of Fermat's own
infinite descent method...

(2016-01-30) Power Tools
The worth of some general theorems is in the proofs of other theorems.

In mathematics, any proven result can be called a theorem.
However, that name is best reserved to general results which give rise to
interesting proving techniques. Here are a few examples:

Pigeonhole principle (Dirichlet, 1834):
If there are fewer drawers than items stored in them, there is a drawer with several items in it.

(2018-09-22) Treacherous Patterns
Some obvious generalizations may not be true...

Inspired by Greg Egan
(2018-09-16)
John Baez remarked
(2018-09-21)
that the following equality holds when n is below
9.8 10^{42} but surely fails when it's above 7.4 10^{43}.

Greg soon proved that this equality holds if and only if:

n < 15341178777673149429167740440969249338310889 = 1.534... 10^{43}

(2019-05-17 13:00 PDT) A botched proof of
Fermat's last theorem.
Fermat himself most likely had in mind an invalid argument like this.

Famous open problems have always attracted the attention
of people poorly equipped to tackle them. Well, the very fact that a problem remains
open long enough to become famous shows that almost nobody is up to it!

If you ever find an easy solution to something which has resisted
the attacks of the best mathematicians for decades (or centuries)
you can be dead sure that you've made a mistake somewhere.
Problems with easy solutions get solved before they become famous.

Legend has it that a prestigious mathematics department once entrusted
graduate students to fill out a pre-printed form in reply to proposed proofs
of Fermat's last theorem, just pointing out the location of the first mistake.

As I received one such claim again in print
a couple of days ago. It looked at first like a classical mistake.
I was wrong (it was something even more trivial) but the mistake I had
in mind is interesting enough to discuss as an example of
what not to do. Here we go:

We only have to consider odd values of the exponent n, since a solution
for any other exponent beyond 2 would translate into a solution either for exponent 4
(a well-known impossibility)
or for such an odd exponent n :

c^{ n} = a^{ n} + b^{ n}
(a > 0, b > 0, n ≥ 3 is odd)

In this, c^{ n} must be divisible by s = a+b > 1
since, for any odd n :

Let x = a + b - c.
We have c = s - x
and, by the binomial theorem :

c^{ n} =
( s - x )^{n} =

n

å

i = 0

(

n i

)

s^{ n-i} (-x)^{i}

As this sum and its first terms are divisible by s, so is the last term (i = n).
That means that x^{ n} is divisible by s > 1.
(So, x and s aren't coprime.)

Let w be the greatest common divisor of x and s,
which is to say that we have two coprime positive
integers u and v such that:

x = w u and s = w v
(where u < v since x < s).

As we knew both sides of our last equation to be divisible by v w,
we may cancel one w and obtain an equation
whose two sides are divisible by v:

w^{ n-1} ( v - u )^{n} = w^{ n-1}

n

å

i = 0

(

n i

)

v^{ n-i} (-u)^{i}

The last term of the rightmost sum is -u^{ n},
which is coprime with v because u is. As v
divides its other terms, that sum is coprime with v.

Therefore, the overall divisibility by v (only)
proves that v divides w^{n-1}.

The mistake would have been to ignore that possibility
and carelessly cancel all factors of w
leading to a contradiction and a fake "proof".