1. The target: a precise abc inequality

Take positive integers $a,b,c$ with $a+b=c$ and $\gcd(a,b)=1$. The other pairwise gcds are then also 1: $\gcd(a,c)=\gcd(a,a+b)=1$ and similarly $\gcd(b,c)=1$. For any positive integer $n$, define its radical

\[ \operatorname{rad}(n)=\prod_{\substack{p\text{ prime}\\p\mid n}}p, \qquad \operatorname{rad}(1)=1. \]

Write $R=\operatorname{rad}(abc)$. The usual abc conjecture says

\[ \forall\varepsilon>0\ \exists K_\varepsilon>0\ \forall(a,b,c)\quad c\leq K_\varepsilon R^{1+\varepsilon}. \tag{ABC} \]

The quantifiers matter: one constant works for all primitive positive triples at a fixed $\varepsilon$; it may depend on $\varepsilon$, and this statement does not give a way to compute it. The alternative symmetric formulation uses coprime nonzero $A+B+C=0$, with $\max(|A|,|B|,|C|)$ in place of $c$ (Goldfeld, §1, PDF p. 1).

Why count primes instead of their powers?

For $n=\prod p^{v_p(n)}$, where $v_p(n)$ is the exponent of $p$,

\[ \log n=\sum_p v_p(n)\log p,\qquad \log\operatorname{rad}(n)=\sum_{v_p(n)>0}\log p. \]

Thus the radical remembers which primes occur and forgets how many times they occur. For $1+8=9$, the product is $2^3 3^2$, so $R=2\cdot3=6$, not $72$. The quality $q(a,b,c)=\log c/\log R$ is about $1.226$. A sharper example is $3+125=128$: $R=2\cdot3\cdot5=30$, and $q\approx1.427$. These finite examples do not disprove (ABC); the conjecture permits exceptions to every fixed exponent $1+\varepsilon$.

Taking logarithms expresses the proposed bound as $\log c\leq(1+\varepsilon)\log R+\log K_\varepsilon$. This is a useful target shape, not permission to substitute another theory's notion of "volume" for $\log R$ without proof.

Why "finitely many exceptions" means the same thing

A frequently used equivalent formulation is: for every $\eta>0$ only finitely many primitive positive triples satisfy $c>R^{1+\eta}$.

To get (ABC) from this form, fix $\varepsilon$. Outside its finite exceptional set, $c\leq R^{1+\varepsilon}$; choose $K_\varepsilon$ larger than 1 and all the finitely many ratios $c/R^{1+\varepsilon}$ in that set.

Conversely, assume (ABC) holds for every positive exponent. Given $\eta>0$, apply it with $\varepsilon=\eta/2$. If $c>R^{1+\eta}$, then $R^{\eta/2}<K_{\eta/2}$, so $R$ is bounded. The same uniform inequality now bounds $c$; hence there are only finitely many possible positive triples. Taking a smaller exponent is essential: (ABC) with exponent $\eta$ alone does not supply this argument.

Why the positive $\varepsilon$ is necessary. The same bound with exponent exactly 1 and one universal constant is false. Following Conrad's exercise 3, let $a=9^{2^k}-1$, $b=1$, and $c=9^{2^k}$ for $k\geq0$. Induction gives $v_2(a)=k+3$: the base case is $9-1=8$; at each step $x=9^{2^k}\equiv1\pmod8$, so $v_2(x^2-1)=v_2(x-1)+v_2(x+1)=v_2(x-1)+1$. The radical retains just one of these $k+3$ factors of 2. Since $\operatorname{rad}(c)=3$, $$ R=3\operatorname{rad}(a)\leq\frac{3a}{2^{k+2}}, \qquad \frac cR>\frac{2^{k+2}}3\longrightarrow\infty. $$ Thus no fixed $K$ can make $c\leq KR$ hold for every triple. This family alone says nothing against the conjectured bounds with positive $\varepsilon$.

A consequence of the quantifiers, not a proof of abc. If primitive positive integers satisfied $x^n+y^n=z^n$, apply (ABC) with $\varepsilon=1/3$ to $a=x^n,b=y^n,c=z^n$. Since $R=\operatorname{rad}(xyz)\leq xyz<z^3$, we would get $z^n\leq K_{1/3}z^4$. For $n>4$ and $z\geq2$, this forces $2^{n-4}\leq K_{1/3}$: the same constant rules out solutions for all sufficiently large exponents. This illustrates why "one constant for every triple" is stronger than checking examples.

An elementary view of the elliptic-curve bridge

For an elliptic curve $E/\mathbb Q$, the minimal discriminant $\Delta_E$ measures degeneration with multiplicities, while the conductor $N_E$ records the bad primes with reduction-dependent exponents. The Szpiro-type estimate in Goldfeld, §4, PDF p. 7 has the conjectural shape $$ |\Delta_E|\leq C_\delta N_E^{6+\delta}\quad(\delta>0). $$

For a primitive abc triple, the associated Frey--Hellegouarch curve is $$ E_{a,b}:\quad y^2=x(x-a)(x+b). $$ Its cubic roots $0,a,-b$ have pairwise differences of absolute values $a,b,c$, so this model's discriminant is $16a^2b^2c^2$. For $1+8=9$, the three roots of the cubic are $0,1,-8$. The displayed integral model has discriminant $16(1\cdot8\cdot9)^2=82944=2^{10}3^4$, while its abc radical is $R=6$. This calculation explains why discriminant exponents and prime support differ; it does not compute the minimal discriminant or conductor.

Goldfeld explains that after passing to a minimal model its discriminant is $(abc)^2$ times a bounded power of 2, and its conductor is $R$ times a bounded power of 2. Conditionally on the Szpiro-type estimate, this yields $$ (abc)^2\ \ll_\delta\ R^{6+\delta} \quad\Longrightarrow\quad (abc)^{1/3}\ \ll_\eta\ R^{1+\eta} \quad(\eta=\delta/6). $$

Goldfeld calls this last inequality weak abc (§1, PDF p. 1). The usual (ABC) immediately implies weak abc because $(abc)^{1/3}\leq c$; the reverse implication is not derived here. For instance, when $a=1,b=c-1$, the geometric mean $(abc)^{1/3}$ grows like $c^{2/3}$, not $c$: a bound on that mean does not by itself give the desired bound on $c$ with the same exponent. Do not turn this schematic conditional bridge into "Szpiro has been proved," or silently identify the weak inequality with the target (ABC). Establishing which precise inequality IUT IV claims is a separate task.

Check your understanding

For $3+125=128$, why does the factor $5^3$ contribute only $5$ to $R$? Because the radical discards the exponent $v_5(125)=3$.
Why does a constant in (ABC) not rule out the example $3+125=128$? Because it can absorb any finite set of high-quality triples.
In the finite-exceptions argument, why use $\eta/2$ rather than $\eta$? It forces $R$ to be bounded before bounding $c$.

Further elementary exercises: Keith Conrad, Analogies between Z and F[T], homework 5, PDF p. 1. This chapter verifies the target and elementary implications, not IUT.