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A 2-adic integer is an element of the p-adic integers for $p=2$, equivalently a compatible sequence of residues modulo $2^r$. Its binary p-adic digits begin with its parity. The highlighted residue-tree path for $13$ begins $1,0,1,1$, while modular inversion converges to a p-adic inverse computes the infinite expansion of $1/3$.

A 7-adic integer is an element of the p-adic integers for $p=7$, equivalently a compatible sequence of residues modulo $7^r$. Its p-adic digits lie in $\{0,1,\dots ,6\}$. For example, $10=3+1\cdot 7$ has low-to-high digits $3,1,0,\dots$.

For $n\ge 0$, define $B_0(X)=1$ and $B_n(X)=X(X-1)\cdots (X-n+1)/n!$. The binomial coefficient formula shows $B_n(m)\in \mathbb{Z}$ for every nonnegative integer $m$. Since these integers are dense in the p-adic integers, continuity of the polynomial implies $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$, so it is an integer-valued polynomial there and $\Vert B_n{\Vert }_{\infty }=1$. The Pascal's identity gives $\Delta B_{n+1}=B_n$ for the forward difference operator.

Every unit $\overline{f}\in {\mathbb{F}}_p[[T]{]}^{\times }$ has a unique Coleman norm operator-fixed lift to ${\mathbb{Z}}_p[[T]{]}^{\times }$. Choose any lift $f$ and use $Nf\equiv f(\mathrm{mod}p)$. The consecutive quotients of the iterates are $N^n(Nf/f)\in 1+p^{n+1}R$ by Coleman norm contraction, so they converge coefficientwise uniformly in the p-adic integers. Their limit is norm-fixed by continuity. Two such lifts have norm-fixed quotient in $1+pR$ and therefore coincide. Uniqueness also makes the lifting map multiplicative and independent of the initial lift.

The continuous functions from the p-adic integers to the P-adic numbers form a Banach space over ${\mathbb{Q}}_p$ with supremum norm $\Vert f{\Vert }_{\infty }={\sup }_{x\in {\mathbb{Z}}_p}|f(x){|}_p$. Compactness of the domain bounds the norm, and uniform convergence in the complete codomain proves completeness. Translation and the forward difference operator $\Delta f(x)=f(x+1)-f(x)$ are bounded operators, with $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$.

In the ordinary group algebra of the additive p-adic integers, let $I$ be the augmentation ideal. The kernel of the map to the group algebra of ${\mathbb{Z}}_p/p^r{\mathbb{Z}}_p$ is generated by $[p^{r}h]-1=([h]-1{)}^{p^r}$, so it is contained in $I^{p^r}$. The quotient is the cyclic-group algebra ${\mathbb{F}}_p[T]/(T^{p^r})$, so its augmentation ideal has zero $p^r$th power and the reverse inclusion holds. Every finite degree of the associated graded ring can be read from such a quotient, giving ${\mathrm{gr}}_I{\mathbb{F}}_p[{\mathbb{Z}}_p]\cong {\mathbb{F}}_p[X]$. The completion is ${\mathbb{F}}_p[[T]]$, but completion is not needed for the graded calculation.

Two p-adic integers agree in their first $n$ low-order p-adic digits precisely when their difference is divisible by $p^n$. Equivalently,If the first differing digit has index $j$, the remaining leading coefficient is nonzero modulo $p$, giving exact p-adic valuation $j$ and p-adic metric $p^{-j}$. If no digit differs, the two elements are equal and their distance is zero. Truncating at $r_n$ therefore gives $|x-r_n{|}_p\le p^{-n}$, with a stricter bound possible when further digits are zero.

The Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

Hensel lifting improves a solution of a polynomial congruence modulo a prime power to one modulo a higher power. For $f\in \mathbb{Z}[X]$, suppose $f(a)\equiv 0(\mathrm{mod}{p}^k)$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}p)$. A lift $a+t{p}^k$ satisfiesThe invertible derivative therefore determines a unique $t$ modulo $p$.

For $f(X)=X^2-2$, the root $3$ modulo seven lifts to $10$ modulo $49$, since $3+7=10$ and $10^2-2=98$. Repetition gives a root in the p-adic integers. The simple-root hypothesis ensures uniqueness; when the derivative vanishes modulo $p$, there may be multiple lifts or none.

Modulo three the right side is always minus one, a nonsquare. At two, $x=-2$ gives the square 2-adic integer $-7\equiv 1(\mathrm{mod}8)$. At twenty-three the same value is a nonzero square modulo the prime and lifts by Hensel's lemma. At every other prime at least five, the model has good reduction of an elliptic curve and the Hasse bound gives more than its single point at infinity, hence an affine smooth point. Lifting that point gives p-adic integer coordinates.

For the Mahler coefficients of a continuous function on the p-adic integers, finite binomial inversion gives $c_n={\sum }_{j=0}^n(-1{)}^{n-j}(\genfrac{}{}{0ex}{}{n}{j})f(j)$. Equivalently their exponential generating function is the displayed identity in the formal power-series ring ${\mathbb{Q}}_p[[T]]$. Each coefficient involves only finitely many terms; no claim of convergence for every $p$-adic value of $T$ is needed.

Let $I$ be a complete discrete valuation ring with a continuous ${\mathbb{Z}}_p$-algebra structure. Every continuous function on ${\mathbb{Z}}_p$ with values in $I$ has a unique uniformly convergent expansion in the binomial polynomials, with $a_n={\Delta }^{n}f(0)$. This is the coefficient-ring version of the Mahler theorem. To see that $a_n\to 0$, approximate the function modulo any power of the maximal ideal by a function of period $p^k$. In the expansion of $(E-1{)}^{p^k}$, all intermediate binomial coefficients are divisible by $p$, while $(E^{p^k}-1)$ kills the periodic function. Iterating makes its differences vanish modulo the prescribed ideal power. Finite differences do not increase the uniform norm, so the same conclusion holds for the original function. Binomial inversion gives agreement at the nonnegative integers, and their density in the p-adic integers gives the expansion everywhere. This gives the dual description of p-adic measures by arbitrary bounded sequences of binomial moments, and hence the Amice transform.

The cosets $a+p^n{\mathbb{Z}}_p$ form a neighborhood basis in the additive group of p-adic integers. An open subgroup contains $p^n{\mathbb{Z}}_p$ for some $n$, so has finite index. Conversely, if the index is $m$, the finite quotient is annihilated by $m$, giving $m{\mathbb{Z}}_p\subset H$. Multiplication by the unit part of $m$ is bijective, so $m{\mathbb{Z}}_p=p^{v_p(m)}{\mathbb{Z}}_p$ is open. This proof applies to abstract subgroups without assuming beforehand that they are closed.

The p-adic digits are the unique coefficients in $x={\sum }_{j\ge 0}{b}_j{p}^j$ for a p-adic integer $x$. Taking least nonnegative representatives of its congruence classes,so each new residue determines one more p-adic digit. For $13$ in ${\mathbb{Z}}_2$, the low-to-high digits are $1,0,1,1,0,\dots$, giving $13=1+0\cdot 2+1\cdot 4+1\cdot 8$. A finite prefix describes a whole coset, not a unique infinite expansion.

The p-adic integers retain arithmetic information modulo arbitrarily high powers of one prime number $p$. They form the subringof the P-adic numbers. Their p-adic absolute value bounds the allowed denominators: $1/3$ belongs to ${\mathbb{Z}}_2$, while $1/2$ does not. The name therefore describes a local arithmetic ring, not just the ordinary integers.^[ref]

A concrete description gives an element by a compatible sequence of congruence classes:The inverse limit condition says that gaining more precision must preserve the digits already determined. Each $r_n$ is a residue modulo $p^n$; choosing its least nonnegative integer representative gives $0\le r_n<p^n$. The ordinary integer $13$ gives residues $1,1,5,13$ modulo $2,4,8,16$.

The useful feature is that one exact object organizes an unlimited family of finite modular arithmetic calculations. Addition and multiplication are performed modulo each $p^n$, and the results remain compatible. A p-adic digit expansion makes this concrete:The expansion is unique and converges in the p-adic metric; its higher powers have smaller p-adic absolute value. The digit agreement and p-adic distance criterion tells exactly which finite digits a precision bound preserves.^[ref]

The p-adic metric on the P-adic numbers is induced by the p-adic absolute value. Its ultrametric inequality is $d_p(x,z)\le \max (d_p(x,y),d_p(y,z))$. On p-adic integers, digit agreement and p-adic distance expresses metric precision as shared low-order digits.

For a prime number $p$, a p-adic number is an element of the field ${\mathbb{Q}}_p$ obtained by completing the rational numbers for the metric $d_p(x,y)=|x-y{|}_p$. Here $|\cdot {|}_p$ is the p-adic absolute value. The Cauchy-sequence construction of a metric completion supplies limits for every Cauchy sequence, and addition, multiplication, and the p-adic absolute value extend to this complete metric space. The rational numbers form a dense subset. ^[ref]

For a concrete example with $p=2$, consider the partial sums of a geometric series:The first values $1,3,7,15,31$ have distances $1/2,1/4,1/8,1/16,1/32$ from $-1$ in the 2-adic absolute value. ConsequentlyThis is a statement about convergence in a metric space with the 2-adic absolute value. In the ordinary real number absolute value, the same partial sums grow without bound. Choosing a metric changes which approximation errors shrink.

More generally every nonzero P-adic number has a unique digit expansionOnly finitely many negative powers occur, while positive powers can continue indefinitely. Written with the highest powers on the left, the digits extend infinitely to the left; for example the P-adic number $-1$ at $p=2$ is written $\dots 111_2$. The p-adic integers are exactly the elements with $|x{|}_p\le 1$, whose digit expansions use only nonnegative powers. ^[ref]

The construction can be understood through modular arithmetic. Successively choose the low-order digits to agree modulo $p$, $p^2$, and higher powers. After fixing digits below $p^n$, the remaining error has p-adic absolute value at most $p^{-n}$; the partial sums form a Cauchy sequence, so completeness supplies a limit. If two expansions first differ at position $j$, their difference has p-adic valuation $j$, proving uniqueness. This turns compatible finite congruence classes into one exact number. In particular, Hensel's lemma upgrades suitable approximate roots of a polynomial modulo powers of $p$ into roots in the p-adic integers.

A p-adic unit is an invertible element of the ring ${\mathbb{Z}}_p$. For a p-adic integer $x$, this is equivalent to $v_p(x)=0$, $|x{|}_p=1$, or a nonzero initial p-adic digit $b_0$. A rational number $a/b$ is a p-adic unit when neither its reduced numerator nor its denominator is divisible by $p$.

Thus $3$ and $1/3$ are p-adic units at $p=2$, while $2$ is a p-adic integer with no inverse inside ${\mathbb{Z}}_2$. The modular inversion converges to a p-adic inverse construction computes the inverse of an integer unit to arbitrary finite precision.

For a short Weierstrass equation of an elliptic curve with coefficients in the p-adic integers, a point with nonintegral $x$ has valuations $v_p(x)=-2m$ and $v_p(y)=-3m$ for some integer $m\ge 1$: the term $x^3$ strictly dominates $Ax+B$ in the equation, so $2v_p(y)=3v_p(x)$. Its formal parameter $T=-x/y$ therefore has valuation $m\ge 1$. These points, together with the identity, form the formal neighborhood $E_1({\mathbb{Q}}_p)$ of the identity, with parameter in $p{\mathbb{Z}}_p$ and the elliptic formal group law over the p-adic integers. For a point with $x\in {\mathbb{Z}}_p$, the equation gives $y\in {\mathbb{Z}}_p$ too.

The formal neighborhood has no nontrivial torsion. Multiplication by an integer coprime to $p$ has series $[m](T)=mT+O(T^2)$ and preserves $|T{|}_p$, so any torsion there must have $p$-power order. If $p\ge 3$, part (i) gives $|T{|}_p\ge p^{-1/(p-1)}>p^{-1}$, contradicting $T\in p{\mathbb{Z}}_p$.

The printed hypotheses also allow $p=2$. In that case, a nontrivial torsion point in the formal neighborhood would have order a power of two and hence yield a point of order two in that neighborhood. But every nonidentity point of order two on a short Weierstrass equation of an elliptic curve has $y=0$ and $x$ a root of the monic cubic over the p-adic integers $X^3+AX+B$. Such a root lies in the p-adic integers, so it cannot belong to the formal neighborhood, whose nonidentity points have nonintegral coordinates. This excludes the remaining case. This is the torsion-free formal subgroup for a short Weierstrass equation. Thus every affine torsion point satisfiesWith good reduction, the reduction homomorphism $E({\mathbb{Q}}_p)\to \tilde{E}({\mathbb{F}}_p)$ has kernel exactly this torsion-free formal neighborhood. Its restriction to the torsion subgroup is therefore injective, proving the required subgroup isomorphism. The short model over the p-adic integers is essential to the prime-two argument; no blanket torsion-injectivity assertion for arbitrary models is needed.

The three suggested values areThere is an obstruction at $p=3$: every integer residue satisfies $x^3-x\equiv 0(\mathrm{mod}3)$, whereas $-1\equiv 2$ is not a square modulo three. Reduction of a solution in the p-adic integers would give an impossible solution modulo three.

At $p=2$, take $x=-2$. The odd 2-adic unit in a ring $-7$ is congruent to one modulo eight, so it is a square in ${\mathbb{Z}}_2$. We use the standard 2-adic square criterion that an odd unit in a ring is a square exactly when it is one modulo eight. Its sufficiency follows by successively adjusting an odd square root modulo $2^m$ by $2^{m-1}$ to obtain the next required square congruence. Thus $y^2=-7$ has a 2-adic integer solution.

For $p\ge 5$ other than $23$, the discriminant of $y^2=x^3-x-1$ isso this model has good reduction of an elliptic curve. Hasse's theorem on elliptic curves givesSince there is just one point at infinity, there is an affine point. Smoothness means that at least one partial derivative of $y^2-x^3+x+1$ is a unit in a ring modulo $p$, so Hensel's lemma lifts the affine point to ${\mathbb{Z}}_p^2$. The last exceptional odd prime is also soluble: at $p=23$, $h(-2)=-7\equiv 16$ has square roots $\pm 4$, which are simple roots of $Y^2+7$ modulo $23$, and lift by Hensel's lemma.

For small-prime checks the suggested values also give $x=0$ at $p=5$ and $x=3$ at $p=7$, where respectively $-1\equiv 2^2$ and $23\equiv 3^2$. ThereforeThe same prime list holds if one only asks for a solution modulo $p$, but the lifting above establishes the asserted p-adic integer solutions. This is local solubility of the cubic x cubed minus x minus one.