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I was comparing the statement of the ABC conjecture on Wikipedia to an article I found geared towards laymen. The propositions differ slightly, but they would be equivalent if the following were true:

If $a$ and $b$ are coprime, and $a + b = c$, is $c$ coprime to both $a$ and $b$?

By division algorithm, if (without loss of generality) $a > b$, then $a$ and $c$ must be coprime: the first iteration of the algorithm subtracts $c = 1*a + b$, and then you're left with $a$ and $b$, which are coprime by proposition and so generate 1 as the final result.

If $b$ is prime, I can imagine a proof by contradiction via the division algorithm.

For more complicated $b$ I am not sure it holds.

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up vote 4 down vote accepted

$$ (2\cdot3\cdot13) + (5\cdot7\cdot11) $$ If the whole thing were divisible by $2$, then since the first term is divisible by $2$, so would be the second term. And similarly for the other primes involved, and similarly if the roles of the two terms are reversed. So the sum cannot be divisible by any of the primes involved. Therefore other primes besides those must exist. So this is another way to prove the infinitude of primes, PROVIDED you know about uniqueness of prime factorizations, which is tacitly relied on above.

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$4$ is coprime to $7$, and their sum $11$ is coprime to both.

On the other hand, if $p\mid a$ and $p\mid a+b$, then $p\mid (a+b)-a=b$.

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Pythagorean squares are always the sum of co-prime numbers

$$3^2+4^2 = 5^2$$ $$9 + 16 = 25$$ $$3\cdot 3 + 4\cdot 4 = 5\cdot 5$$ $$\text{--- or ---}$$ $$8^2 + 15^2$$ $$64 + 225 = 289$$ $$2\cdot 2\cdot 2 + 3\cdot 5 = 17\cdot 17$$

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