I have a question about the last step in the proof of Mason's Theorem. I will write it.

Mason's Theorem. If $a, b$ and $c$ are relatively prime polynomials such that $a + b = c$, then $$\max \{ \deg(a) , \deg(b) , \deg(c)\} \leq N_0(abc) -1,$$ where $N_0(f)$ is the number of distinct factors of $f$.

Proof: Let $\displaystyle a(t) = \prod_{i=1}^{s_a} (t-\alpha_i)^{u_i},~ b(t) = \prod_{i=1}^{s_b}(t-\beta_i)^{v_i}, ~c(t) = \prod_{i=1}^{s_c}(t-\gamma_i)^{w_i} $

$N_0(abc) = s_a + s_b + s_c$. Let $\displaystyle f = \frac{a}{c},~ g= \frac{b}{c}$. Hence $f + g = 1 \Rightarrow f' + g' = 0 \Rightarrow f' = - g'$.

$\displaystyle\frac{f'}{f} = \sum_{i=1}^{s_a} \frac{u_i}{t-\alpha_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}, ~~ \frac{g'}{g} =\sum_{i=1}^{s_b} \frac{v_i}{t-\beta_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}$

$\displaystyle \frac{b}{a} = \frac{g}{f} =-\frac{f'/f}{g'/g}=-\frac{M f'/f}{M g'/g},\tag{1} $

where $M =\prod_{i=1}^{s_a}(t-\alpha_i) \prod_{i=1}^{s_b}(t-\beta_i)\prod_{i=1}^{s_c}(t-\gamma_i)$.

So the denominator and nominator in $(1)$ has the degree at most $s_a + s_b + s_c -1 $. Then we conclude the inequality of the theorem.

I understand all the proof except the last thing.

  • 2
    $\begingroup$ Perhaps you want see What is the abc conjecture?, by UConn Math, from YouTube, around 33'. It isn't neccesary a response of this comment, it is only you want see it. $\endgroup$
    – user243301
    May 9, 2016 at 17:00

1 Answer 1


Assume that polynomials have coefficients in a field of characteristic $0$. From $b/a=-(f^{'}/f)/(g^{'}/g)$ we observe that $\deg(f^{'}/f)=-\infty$ if $b=0$ and $\deg(f^{'}/f)=-1$ if $a$ equals to zero (I edit here, because $f$ is defined as $a/c$ and $\deg(f^{'})=\deg(f)-1$).

Hence $$\deg(M f^{'}/{f})\leq n_{0}(abc)-1$$

thus from $-a(M\cdot f^{'}/{f})=b(M\cdot g^{'}/{g})$, we deduce that $a$ divides $M\cdot \frac{g^{'}}{g}$ since $\gcd(a,b)=1$, this and previous divisibility relation implies $\deg(a)\leq n_{0}(abc)-1$.

A similar argument works for $\deg(b)\leq n_{0}(abc)-1$.

Obviously from $a+b=c$ we have $\deg(c)\leq\max(\deg(a),\deg(b))$. Putting together these results we obtain $\max(\deg(a,b,c))\leq n_{0}(abc)-1$.

References (I haven't an open access, I took the details from this, I hope that it is possible):

Jeffrey Paul Wheeler' Thesis, The abc conjecture.


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