ABC Con Gatcha

For all triples \((a,\, b,\, c)\) of coprime positive integers, with \(a + b = c\),

we see that for \(\varepsilon\gt 0\) where

\(c\gt rad(abc)^{1+\varepsilon}\)

with the radical of a positive integer \(n\), the radical of  \(n\), denoted \(rad(n)\), is the product of the distinct prime factors of \(n\);

is the small blue circle smaller than the red circle of radius \(a+b=c\).

The diagram does not count the valid number of triples for which the blue circle is small, but if we consider all valid triples of co-prime and scale each corresponding green circle \(\small{abc}\) to a uniform size, and allow all with smaller blue circle to sieve through the red circle, then obviously, by the  pigeonhole principle, since the green circles are bigger and the blue circle can be as big as the green circle, only some blue circles will fall through the red circle at the center.

Is the number of blue circles that fell through finite?

We note that \(abc\gt \sqrt{ab^2+a^2b}\gt c \gt rad(abc)^{1+\varepsilon}\), in other words, the blue circle is the smallest and has radius less than \(\sqrt{abc}\).

Using Euler's totient function,

\(\varphi(ab)=\varphi(a)\varphi(b)\)

where \(a\) and \(b\) are co-prime.  We may,

\(\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\)

where \(a\), \(b\) and \(c\) are co-prime.  As,  \(n\gt\varphi(n)\)

\(abc\gt\varphi(abc)\)

and

\(abc\gt c\)

So, we have either,

\(abc\gt\varphi(abc)\gt c\gt\varphi(c)\)

or

\(abc\gt c\gt \varphi(abc)\gt\varphi(c)\)

because,

\(\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\gt\varphi(c)\)

When,

\(abc\gt\varphi(abc)\gt c\gt rad(abc)^{1+\varepsilon}\)

Since, \(abc\) is not prime, \(\varphi(abc)\) is finite; that the number of positive integer not greater than \(abc\), and relatively prime with \(abc\) is countable, the number of blue circles that fell through is also countable.  If \(abc\) is prime (which is impossible as it has three factors) than \(\varphi(abc)=abc-1\), which is as large as \(abc\).

When

\(abc\gt c\gt \varphi(abc)\gt rad(abc)^{1+\varepsilon}\)

which is the same as the first case, the position of \(c\) is irrelevant.  In both cases, however,  the count of terms \(\varphi(abc)\) is greater than the product of terms \(rad(abc)^{1+\varepsilon}\), which can be true (in the limiting case of equality \(\varphi(abc)=rad(abc)^{1+\varepsilon}\) ) only for the first two prime numbers \(1\) and \(2\).

When

\(abc\gt c\gt rad(abc)^{1+\varepsilon}\gt \varphi(abc)\gt\varphi(c)\)

we have,

\(c\gt rad(abc)^{1+\varepsilon}\gt \varphi(c)\) --- (*)

where \(rad(abc)^{1+\varepsilon}\) is squeeze between \(c\) (\(c\gt a\) and \(c\gt b\)) and \(\varphi(c)\) which makes \(rad(abc)^{1+\varepsilon}\) countable.

For expression (*) to be possible, when \(c\) is large but not divisible by large powers of prime then, as \(rad(abc)^{1+\varepsilon}\) exists in the range from \(c\) to \(\varphi(c)\), this implies that \(a\) and \(b\) must be of high powers of primes, because \(a\), \(b\) and \(c\) are co-prime and factor \(a\) and \(b\) must contribute less to \(rad(abc)\) than \(c\), on condition that \(rad(abc)\lt c\) is true for the triple \((a,\, b,\, c)\).

This is not to be taken too seriously...