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The figure illustrates Pierre de Fermat's use of adequality to construct a tangent to a parabola. Reading the local exam image is decisive: $D$ is the vertex of a parabola, $DC$ lies on its axis of symmetry, $B$ lies on the parabola, and the proposed tangent $BE$ meets the extended axis at $E$. The horizontal through the auxiliary point $O$ meets the axis at $I$; $O$ lies on $BE$, outside the parabola. In particular, $O$ must not be treated as a second point of the parabola. The target is the subtangent $CE$. The diagram and this construction occur in the tangent discussion associated with Varia opera mathematica by Pierre de Fermat. The 1679 book was posthumous: its date does not date the invention of the method.^[ref][ref]

Write $d=CD>0$, $a=CE>0$, and $e=CI$, with $I$ between $C$ and $D$. These are lengths, and the increment $e$ is distinct from the labelled point $E$. Then $DI=d-e$ and $EI=a-e$. The parabola's characteristic relation makes the square of an ordinate proportional to its distance from the vertex of a parabola. At height $I$, the actual ordinate of the parabola is shorter than $OI$. Hencewhere the equality follows from similar triangles $BCE$ and $OIE$.

The method of adequality replaces the comparison by a relation used to extract the condition of contact:Expanding and cancelling $d{a}^2$ gives $-2dae+d{e}^2\sim -a^{2}e$. Divide by the nonzero increment $e$, then suppress the remaining term containing $e$. The result is $a^2=2da$, so, since $a>0$,Thus $DE=DC$: the intersection with the axis lies as far above the vertex of a parabola as $C$ lies below it. This is a concrete construction, not just a rule for computing a numerical slope.

A modern check clarifies why the calculation succeeds without pretending that Pierre de Fermat used a modern limit. Put $D=(0,0)$, let $y$ increase down the axis, and write the parabola as $x^2=ky$, $k>0$. For $B=(b,d)$, with $b^2=kd$, the proposed tangent through $E=(0,-d)$ is $y=2bx/k-d$. Substitution givesIt meets the parabola with a double root at $B$, and for every other point its $y$ coordinate is less than that of the parabola. This also verifies the strict comparison used for $O$.

The significance is a systematic algebraic condition for geometric contact. Subtracting neighbouring expressions, dividing by an increment, and discarding its remaining terms anticipates differentiation. But the historical question is precisely what licenses the last step. The symbol $\sim$ above marks adequality, not ordinary equality at a finite displacement. Treating the construction as an already stated epsilon-delta definition of a limit would conceal the methodological innovation and its unresolved explanatory burden.

Since $26\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{26}]$ with discriminant $104$. Write $P=(2,ϵ+1)=(2,\sqrt{26})$ and $Q_{\pm }=(5,ϵ\pm 1)$. Reduction of $X^2-26$ modulo two gives a double root, and modulo five gives the roots $\pm 1$. Thus $N(P)=2$, $N(Q_{\pm })=5$, and $(5)=Q_{+}{Q}_{-}$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot 4=2$. The element $ϵ+1=6+\sqrt{26}$ belongs to $P$ and $Q_{+}$ and has field norm ten, so equality of ideal norms givesThe Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every ideal class contains an integral ideal of ideal norm at most five. Three is inert and gives no prime ideal of ideal norm three; ideals of ideal norm four are $P^2$, principal. Thus the class group is generated by $P,Q_{+},Q_{-}$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have field norm $\pm 2$, but $x^2-26y^2=\pm 2$ would give $x^2\equiv 2$ or $11(\mathrm{mod}13)$, neither a quadratic residue. Hence the class number is two.

The unit $ϵ=5+\sqrt{26}$ has field norm $-1$. To show it is fundamental, suppose $1<u<ϵ$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm 1/u$, so $y=(u-u^{\prime })/(2\sqrt{26})$ is a positive integer less than two. Thus $y=1$, and $x^2-26=\pm 1$ forces $x=\pm 5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of ${ϵ}^{-1}$ puts it in $[1,ϵ)$, proving that all units are $\pm {ϵ}^n$.

An element of field norm $\pm 10$ has principal ideal $P{Q}_{+}$ or $P{Q}_{-}$ by prime ideal factorization. The first is generated by $ϵ+1$; the second by $ϵ-1=4+\sqrt{26}$, of field norm $-10$. Two generators of the same principal ideal differ by a unit. Therefore all, and only, the integer solutions areThe two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two field norm signs.

The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

Suppose a perturbed polynomial has the local form $A(x-x_0{)}^2+ϵB+O((x-x_0{)}^3,ϵ(x-x_0),{ϵ}^2)$ with $AB\ne 0$. Dominant balance gives $x-x_0={ϵ}^{1/2}z$ and $A{z}^2+B=0$. The two resulting Puiseux series split a double root; their leading displacements can be real or a complex conjugate pair. Ordinary Taylor series in $ϵ$ cannot describe this splitting.