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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.

Michael Chen
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6 Answers6

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$$ (2\cdot3\cdot13) + (5\cdot7\cdot11) $$ 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.

Michael Hardy
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4

$4$ 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 $p\mid (a+b)-a=b$.

Michael Hardy
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vadim123
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Let $a$ and $b$ be two coprime numbers.We know,in a equation $a+b=c$,if any two of $a,b,c$ are divisible by any other number $x$,then the third one is also divisible by $x$. Now, if $a$ and $b$ are coprime, then the only number to divide both of them is $1$. That means $c$ cannot be divisible by any divisor of either $a$ or $b$. So, $c$ is coprime to both $a$ and $b$.

mlc
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Sajid3176183
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Another proof: $a+b=c$ where $a$ and $b$ are all integers greater than $0$ (as other parameters in the proof) and coprime. Let’s assume $c$ is not coprime with $a$. Then, $p*a’+b=p*c’$ where $p>1.$ Then, following applies: $b=p*(c’-a’)$ This implies that, $b$ also has the factor $p$ which is not possible since it is coprime with $a$. If we repeat same for $b$ and $c$ with an assumed common factor $q$, we reach same conflict and same conclusion. That concludes that, $c$ is also coprime with $a$ and $b$.

nonuser
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C. Kayrak
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Let $a=p_1p_2...p_m$ and $b=q_1q_2...q_n$ where $p_i \ne q_j$, and $p_i,q_j$ are primes.

Then $c=p_1p_2...p_m+q_1q_2...q_n$. Now take any prime factor of either $a$ or $b$ and divide $c$ by it.

WLOG choose $p_1$ so $c/p_1=\frac{p_1p_2...p_m+q_1q_2...q_n}{p_1}=p_2...p_m+\frac{q_1q_2...q_n}{p_1}$. The rightmost fraction cannot be simplified, so $p_1\not\mid c$.

The argument repeats for all $p_i$ and $q_j$, so $c$ is coprime to both $a$ and $b$.

Keith Backman
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-3

Pythagorean squares are always the sum of co-prime numbers

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

epimorphic
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Robin
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