nt.number theory - Heuristically false conjectures

Just run across this question, and am surprised that the first example
that came to mind was not mentioned:

Fermat's "Last Theorem" is heuristically true for $n > 3$,
but heuristically false for $n=3$ which is one of the easier
cases to prove.

if $0 < x leq y < z in (M/2,M]$ then $|x^n + y^n - z^n| < M^n$.
There are about $cM^3$ candidates $(x,y,z)$ in this range
for some $c>0$ (as it happens $c=7/48$), producing values of
$Delta := x^n+y^n-z^n$ spread out on
the interval $(-M^n,M^n)$ according to some fixed distribution
$w_n(r) dr$ on $(-1,1)$ scaled by a factor $M^n$ (i.e.,
for any $r_1,r_2$ with $-1 leq r_1 leq r_2 leq 1$
the fraction of $Delta$ values in $(r_1 M^n, r_2 M^n)$
approaches $int_{r_1}^{r_2} w_n(r) dr$ as $M rightarrow infty$).

This suggests that any given value of $Delta$, such as $0$,
will arise about $c w_n(0) M^{3-n}$ times. Taking $M=2^k=2,4,8,16,ldots$
and summing over positive integers $k$ yields a rapidly divergent sum
for $n<3$, a barely divergent one for $n=3$, and a rapidly convergent
sum for $n>3$.

Specifically, we expect the number of solutions of $x^n+y^n=z^n$
with $z leq M$ to grow as $M^{3-n}$ for $n<3$ (which is true and easy),
to grow as $log M$ for $n=3$ (which is false), and to be finite for $n>3$
(which is true for relatively prime $x,y,z$ and very hard to prove [Faltings]).

More generally, this kind of analysis suggests that for $m geq 3$
the equation $x_1^n + x_2^n + cdots + x_{m-1}^n = x_m^n$
should have lots of solutions for $n<m$,
infinitely but only logarithmically many for $n=m$,
and finitely many for $n>m$. In particular, Euler's conjecture
that there are no solutions for $m=n$ is heuristically false for all $m$.
So far it is known to be false only for $m=4$ and $m=5$.

Generalization in a different direction suggests that any cubic
plane curve $C: P(x,y,z)=0$ should have infinitely many rational points.
This is known to be true for some $C$ and false for others;
and when true the number of points of height up to $M$ grows as
$log^{r/2} M$ for some integer $r>0$ (the rank of the elliptic curve),
which may equal $2$ as the heuristic predicts but doesn't have to.
The rank is predicted by the celebrated conjecture of Birch and
Swinnerton-Dyer, which in effect refines the heuristic by accounting
for the distribution of values of $P(x,y,z)$ not just
"at the archimedean place" (how big is it?) but also "at finite places"
(is $P$ a multiple of $p^e$?).

The same refinement is available for equations in more variables,
such as Euler's generalization of the Fermat equation;
but this does not change the conclusion (except for equations such as
$x_1^4 + 3 x_2^4 + 9 x_3^4 = 27 x_4^4$,
which have no solutions at all for congruence reasons),
though in the borderline case $m=n$ the expected power of $log M$ might rise.

Warning: there are subtler obstructions that may prevent a surface from
having rational points even when the heuristic leads us to expect
plentiful solutions and there are no congruence conditions that
contradict this guess. An example is the Cassels-Guy cubic
$5x^3 + 9y^3 + 10z^3 + 12w^3 = 0$, with no nonzero rational solutions
$(x,y,z,w)$:

Cassels, J.W.S, and Guy, M.J.T.:
On the Hasse principle for cubic surfaces,
Mathematika 13 (1966), 111--120.

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