# 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.

• Duplicate: mathoverflow.net/questions/2779/… – Qiaochu Yuan Nov 26 '10 at 17:35
• The example in your question is a famous example of Selmer. An explanation is given by Keith Conrad here: math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf Note also that the example given below by Qiaochu are reducible. One can show that if $f(x)$ is an integer polynomial in one variable that is irreducible, then it cannot have a solution modulo every prime. The essential point of Selmer's example is that it is irreducible. – Emerton Nov 26 '10 at 18:52
• A proof that there are no solutions over Q is in Cassels' book on elliptic curves. An elementary proof that there are p-adic solutions for all p, using Hensel's lemma and an argument very much in the spirit of Qiaochu's answer (i.e. using a "cube" analogue of the statement that if $a$ and $b$ aren't squares mod $p$ then $ab$ is) can be found at www2.imperial.ac.uk/~buzzard/maths/teaching/10Aut/M4P32/… : I just set it as homework for my students in fact :-) – Kevin Buzzard Nov 26 '10 at 20:30
• Dear A. Pacetti, Here is the argument I had in mind (it is not original to me, though); hopefully I am not butchering it: consider the Galois group $G$ of the splitting field of the polynomial. If the poly. $f$ is irred., then $G$ acts transitively on the roots of $f$, and so group theory shows that (if $f$ has degree $> 1$) then there is a conjugacy of $G$ with no fixed point. The $p$ whose Frobenius elements are equal to this conjugacy class then have the property that $f$ has no root modulo $p$. (Such a root would give a fixed point for the Frobenius of $p$. – Emerton Nov 27 '10 at 1:38
• Another elementary example is $x^2+y^2+z^2+w^2=-1$. Because every positive integer is a sum of 4 squares, this has solutions modulo every $n\ge 2$. – Tim Dokchitser Nov 27 '10 at 11:51

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 $\bmod p$.

• @Qiaochu : CRT ? – Andrés E. Caicedo Nov 26 '10 at 17:43
• I am afraid that this specific example does not have solution modulo 8, but if one multiples by $(x^2-17)$ then it becomes ok. – Fedor Petrov Nov 26 '10 at 17:47
• @Andres: Chinese remainder theorem. @Fedor: oops! I think there is a simpler example along those lines, though. – Qiaochu Yuan Nov 26 '10 at 18:00
• Qiaochu: Out of curiosity, what is Hilbert's family of examples? – Faisal Nov 26 '10 at 19:42
• I approve ! I gave the same example $(x^2-2)(x^2-3)(x^2-6)=0$ thirty-three years ago, at the oral examination of the French teaching contest "Agrégation". – Denis Serre Nov 26 '10 at 19:46

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.

• The example I usually give my students is $x^2-37y^2=3$. The rational solution is $7/2,1/2$ and there's a 2-adic integer solution as well. However it's much harder to prove there's no integer solutions! (I can do it without too much trouble on a computer but...) So I'm very grateful for your example Felipe! – Kevin Buzzard Nov 26 '10 at 20:43
• An easy way to see that there are $3$-adic solutions is to note that $(81/13, 4/13)$ is a solution. – David E Speyer Dec 2 '10 at 18:30
• $(9/4,5/4)$ is a solution too. I should have looked for other rational solutions. – Felipe Voloch Dec 2 '10 at 18:55

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).

• KConrad: it would have been shorter to say "$(1/3,1/3)$ and $(3/5,1/5)$" than "Look for a rational solution....both cases)." :-) – Kevin Buzzard Nov 28 '10 at 0:45
• Yours seems to me like the "best" answer so far though (simplest equation, smallest coefficients). – Kevin Buzzard Nov 28 '10 at 0:45
• A Weierstrass equation in this spirit is y^2 = x^3 - 51 (which BCnrd told me about, and I think he got it from Venkatesh). It has the rational solution (1375/9,50986/27) and it can be solved mod 3^r for any r using x = 1 and for y a square root of -50 mod 3^r (which exists since -50 = 1 mod 3). Thus as a congruence this equation has a solution mod m for any m. That there is no Z-solution follows from Q(sqrt(-51)) having class number 2, which is rel. prime to 3, by the same kind of method used to find the Z-solutions to y^2 = x^3-2. – KConrad Nov 28 '10 at 7:19
• The point of my previous comment was to have a rational point and no integral point, but the rational point doesn't matter. For many nonzero integers $k$ (e.g., $k = \pm 6$) the equation $y^2 = x^3 + k$ has no integral solutions, but for every nonzero integer $k$, the congruence $y^2 \equiv x^3 + k \bmod m$ has a solution for all $m \geq 2$. It suffices by Chinese remainder theorem to show there is a solution when $m$ runs through prime powers, and my answer at mathoverflow.net/questions/134352 does that. (At math.stackexchange.com/questions/875983 I show solns mod $p$.) – KConrad Feb 3 '19 at 2:31
• A simpler example than $y^2 = x^3 - 51$ of such an equation with no integral solution and with a rational solution is $y^2 = x^3 + 11$, with rational solution $(-7/4,19/8)$. That it has no integral solution can be proved with just congruences and knowing when $-1 \equiv \Box \bmod p$, no need for class numbers. See Theorem 2.3 of kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf. The rational solution shows there is a solution mod $p^r$ for odd primes $p$, and to get a solution mod powers of $2$ there is a $2$-adic solution $(x,2)$ where $x^3 = -7$ with $x \equiv 1 \bmod 2$. – KConrad Feb 3 '19 at 5:54

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.

There is an easier example in

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

• Is there a Brauer-Manin obstruction to these examples? – Felipe Voloch Nov 27 '10 at 19:08
• Hi, Felipe. I really would not know. To be more precise, I do not know what the phrase means. But there is such an example with $x^2 + y^2 + z^n$ where $n$ is any odd and composite number. It is known that $x^2 + y^2 + z^3$ integrally represents all integers, and modest evidence suggests the same for any $n$ odd prime. Meanwhile, warranted or not, I think this is related to my first question in January 2010, $2 x^2 + x y + 3 y^2 + z^3 - z,$ answered with some real effort by Kevin Buzzard. There are about 25 of those, one for each discriminant of positive binary forms with class number 3. – Will Jagy Nov 27 '10 at 20:15
• Felipe, the integral Brauer-Manin obstruction has been systematically invstigated by Colliot-Thélène and Xu (math.u-psud.fr/~colliot/CTXuCompositio2009.pdf).  – Chandan Singh Dalawat Nov 28 '10 at 11:00
• @Chandan: Yes, I know about this paper. My question was whether the lack of integral solutions to Will's examples can be explained by an integral Brauer-Manin obstruction. – Felipe Voloch Nov 29 '10 at 18:57
• Felipe and Chandan, thank you for at least discussing this a little. Kap's proof by factorization is in the short paper with the link already given in my answer above. The phenomenon, in my comment above, in positive integral ternary quadratic forms is in a 1995 letter by Kap to J. S. Hsia and Rainer Schulze-Pillot, pdf link attempted here:  zakuski.math.utsa.edu/~kap/Forms/…  and the January 2010 question Kevin answered at:  mathoverflow.net/questions/12486/…  Will  – Will Jagy Nov 29 '10 at 19:51

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.

• This is nice, Christian. – Will Jagy Dec 2 '10 at 20:43

See 6.4.1 in my paper with Rudnick http://www.springerlink.com/content/l1t0071152537186/, 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.