# Solvability of Diophantine equation using congruences where the variables are bounded?

In page 2 of Mordell "Diophantion equations" it is given a necessary condition for the solvability of a Diophantine equation:

Integer solutions of the inhomogeneous equation $f(x) = 0$ can exist only if the congruence $$f(x) = 0 \pmod M$$ has solutions for all integers $M$.

Is the converse true?

More stronger, if $$f(x) = 0 \pmod{p^t}$$ for some $p$ and for every $t$, is it true that $f(x)=0$ is soluble? If not could you please give some examples or references for such problems?

More precisely, if the variables in the (multi-variable) Diophantine equation are bounded, it seems we can apply the converse of the Theorem mentioned in the beginning, i.e., for a given and fixed $X$ $$f(x_1,\ldots,x_s)=0,\qquad |x_i|\leq X$$ it is possible to solve the equation $$f(x_1,\ldots,x_s)=0\pmod{p^t},\qquad |x_i|\leq X$$ for large enough $p^t$, where $p^t$ is at least some constant multiple of $X^d$, where $d$ is the highest degree of $f(x_1,\ldots,x_s)$.

• "Hasse principle" (Google-able) is the principle (sometimes, not always correct) that if a diophantine equation has solutions in p-adic (for all p) and real numbers, then it has rational solutions. This is true for quadratic equations, but not generally. There are failures for certain cubics (maybe Google-able or in Wiki). – paul garrett Aug 31 '16 at 14:07
• The way it is stated in the question (omitting mention of reals), it fails even for quadratic equations: consider $x^2+y^2+z^2+w^2+1$. – Emil Jeřábek Aug 31 '16 at 14:34
• With the added restriction of the fixed bound $X$, the problem becomes trivial. You can compute $M$ so that $|f(x)| \le M$ whenever all $|x_i| \le X$. If prime $p > M \ge |f(x)|$ and $f(x) \equiv 0 \mod p$, then of course $f(x) = 0$. – Robert Israel Aug 31 '16 at 23:54
• For your "more stronger" claim, surely a trivial counterexample is $2x+1=0$, which has a solution modulo $3^t$ for all $t$, but no solution over $\mathbb Z$. – Ilmari Karonen Sep 1 '16 at 0:09

## 1 Answer

There is a vast literature on this subject of counter-examples to the Hasse principle. This local-to-global principle holds for quadratic forms and curves of genus $0$ (more generally for twisted forms of projective spaces), but it fails (sometimes miserably) for curves of genus $1$ and for cubic surfaces, for example. One nice place to get acquainted with these questions is the wonderful article by Mazur

On the passage from local to global in number theory

Author: B. Mazur Journal: Bull. Amer. Math. Soc. 29 (1993), 14-50

http://www.ams.org/journals/bull/1993-29-01/S0273-0979-1993-00414-2/

The most famous example ($3x^3+4y^3+5z^3=0$) is treated by Keith Conrad here : http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi=10.1.1.210.8211