Diophantine equation with no integer solutions, but with solutions modulo every integer It's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo $n$ for every $n$. This fact is stated, for example, in Dummit and Foote (p. 246 of the 3rd edition), where it is also claimed that an example is given by the equation
$$ 3x^3 + 4y^3 + 5z^3 = 0. $$
However, D&F say that it's "extremely hard to verify" that this equation has the desired property, and no reference is given as to where one can find such a verification.
So my question is: Does anyone know of a readable reference that proves this claim (either for the above equation or for others)? I haven't had much luck finding one.
 A: There is an  easier example in 
http://zakuski.math.utsa.edu/~jagy/papers/Experimental_1995.pdf 
where Kap disposed of the concern with the brief "(it is easy to see that the assumption of no congruence obstructions is satisfied)."
The example is, given a positive prime $p \equiv 1 \pmod 4,$ there is no solution in integers $x,y,z$ to
$$  x^2 + y^2 + z^9 = 216 p^3  $$
Robert C. Vaughan wrote to Kap (prior to publication) in appreciation, there was something involved that "could not be detected p-adically." I forget what, it has been years. But we did well, Vaughan got an early draft in time to include the example in the second edition of
The Hardy-Littlewood Method.  
Later for some reason I looked at negative targets, with the same primes I believe it turned out that there were no integer solutions to
$$  x^2 + y^2 + z^9 = -8 p^3.  $$
The significance of the example is not so much as a single Diophantine equation, rather as a Diophantine representation problem in the general vicinity of the Waring problem, but with mixed exponents: given nonegative integer variables $x,y,z$ and exponents $a,b,c \geq 2,$ and given the polynomial $f(x,y,z) =x^a + y^b + z^c,$ if $f(x,y,z)$ represents every positive integer $p$-adically and if $$ \frac{1}{a} +  \frac{1}{b} + \frac{1}{c} > 1,  $$  does  $f(x,y,z)$ integrally represent all sufficiently large integers? The answer is no for the problem as stated, but the counterexamples depend heavily on factorization, and in the end upon composition of binary forms. As this is also the mechanism underlying the simplest examples of  spinor exceptional integers for positive ternary quadratic forms, it is natural to ask whether there is some relatively easy formalism that adds "factorization obstructions" to the well-studied "congruence obstructions." 
See:
http://zakuski.math.utsa.edu/~jagy/Vaughan.pdf
http://en.wikipedia.org/wiki/Waring's_problem
A: An example even easier than Jagy and Kaplansky's
$x^2+y^2+z^9 = 216p^3$, for $p=1 \bmod 4$, is given in:
Sums of two squares and one biquadrate, by R. Dietmann, and C. Elsholtz,
Funct. Approx. Comment. Math. Volume 38, Number 2 (2008), 233-234.
http://www.math.tugraz.at/~elsholtz/WWW/papers/papers26de08.pdf
Here we showed:
$x^2+y^2+z^4=p^2$ has no positive solutions, when $p=7 \bmod 8, p $prime. Once the example is known, it's trivial to prove.
The Jagy-Kaplansky example can be generalized to odd composite exponent, instead of 9. It seems the example above was overlooked for quite a while.
A: Here is another example, which is easy to verify by hand: $x^2+23y^2=41$. Note it has rational solutions (e.g. $(1/3,4/3)$). This provides solutions modulo $m$ if $(m,3)=1$.
For $m$ a power of $3$, there is always a solution with $x=0$. Verifying that it doesn't have integral solutions is trivial.
A: See 6.4.1 in my paper with Rudnick Link, page 62.
The equation is:
$$
-9x^2+2xy+7y^2+2z^2=1.
$$
This equation has a rational solution $(-\frac{1}{2}, \frac{1}{2},1)$, hence it has  solutions modulo $p^n$ for all $p\neq 2$ and all $n$.
In addition, it has a solution $(4,1,1)$ modulo $2^7$, and using Hensel's lemma one can easily check that the equation has solutions modulo $2^n$ for all $n$.
The elementary proof that this equation has no integral solutions is due to Don Zagier and is based on (a supplementary formula to) the quadratic reciprocity law.
A: It is actually quite straightforward to write down examples in one variable where this occurs.  For example, the Diophantine equation $(x^2 - 2)(x^2 - 3)(x^2 - 6) = 0$ has this property: for any prime $p$, at least one of $2, 3, 6$ must be a quadratic residue, so there is a solution $\bmod p$, and by Hensel's lemma (which has to be applied slightly differently when $p = 2$) there is a solution $\bmod p^n$ for any $n$.  We conclude by CRT.  (Edit:  As Fedor says, there are problems at $2$.  We can correct this by using, for example, $(x^2 - 2)(x^2 - 17)(x^2 - 34)$.)
Hilbert wrote down a family of quartics with the same property.  There are no (monic) cubics or quadratics with this property: if a monic polynomial $f(x) \in \mathbb{Z}[x]$ with $\deg f \le 3$ is irreducible over $\mathbb{Z}$ (which is equivalent to not having an integer solution), then by the Frobenius density theorem there are infinitely many primes $p$ such that $f(x)$ is irreducible $\bmod p$.
A: The equation 2x^2 + 7y^2 = 1 has two rational solutions with small relatively prime denominators (hence as a congruence mod m it is solvable for all m by CRT) but it visibly has no integral solutions. Look for a rational solution with denominator 3 and also for one with denominator 5 (small numerators in both cases). 
A: Let us define the size $H$ of a polynomial Diophantine equation $P(x_1,...,x_n)=0$ as: substitute $2$ instead of all variables, absolute values instead of all coefficients, and evaluate. This notion of size has the advantages that (i) there is a finite number of equations with bounded size, so we can do full computer search and find the smallest equation with any given property and (ii) equations with small $H$ really look nice and compact.
Then the smallest Diophantine equation with no integer solutions, but with solutions modulo every integer is the equation
$$
y(x^2+2)=1
$$
with size $H=2(2^2+2)+1 = 13$.
Proof that a solution exists: As usual for polynomial equations, CRT allows us to reduce to the case $n = p^e$ for $p$ prime and $e$ positive. If $p > 2$, then $p^e$ is odd, so $2$ is invertible mod $p^e$. Set $x = 0, y = 2^{-1}$. If $p = 2$, then $3$ is invertible mod $2^e$. Set $x = 1, y = 3^{-1}$.
A: Consider the equation $(2x - 1)(3x - 1) = 0$.  This equation has no integer solutions.  But modulo $n$, it always has a solution.  If $n$ is not a multiple of $2$, we can make $2x -1$ a multiple of $n$.  If $n$ is not a multiple of $3$, we can make $3x - 1$ a multiple of $n$.  Using the Chinese Remainder Theorem, we can handle every other $n$ by piecing together these two solutions.
