# 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). Commented Aug 31, 2016 at 14:07
• The way it is stated in the question (omitting mention of reals), it fails even for quadratic equations: consider $x^2+y^2+z^2+w^2+1$. Commented Aug 31, 2016 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 $f(x) = 0$. Commented Aug 31, 2016 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 $\mathbb Z$. Commented Sep 1, 2016 at 0:09

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
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