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Let $a$ and $b$ be positive integers with no common factors. Then

$a)$ $a+b$ and $a-b$ have no common factor other than $3$ , whatever be $a$ and $b$

$b)$ $a+b$ and $a-b$ have no common factor greater than 2,whatever be $a$ and $b$

$c)$ $a+b$ and $a-b$ have a common factor, whatever be $a$ and $b$

$d)$ none of the foregoing statements is correct.

My approach : I am not proficient in number theory , so my approach till now was pretty lame . I basically included trying to check for the numbers. I first tried to check using consecutive numbers since they are always co primes. However I realised in this case I would be missing out those kind of co-primes both of whose constituents are odd numbers. This is where I hit the roadblock. ( My background is a degree in Electrical Engineering , though with no formal course/training in Number Theory.)

Please tell me the correct approach to solve the question.

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  • $\begingroup$ Try different primes for $a$ and $b$. They are coprime. Say, $a=7$, $b=3$. Then $a+b=10$ and $a-b=4$. So we are (almost) done. $\endgroup$ Apr 5, 2016 at 18:43

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Any factor that $a+b$ and $a-b$ have in common is also a factor of their sum, which is $2a$. And it is a factor of their difference, which is $2b$. If a prime divides $2a$, then either the prime is $2$ or it is a divisor of $a$. And if it divides $2b$, then either it is $2$ or it is a divisor of $b$. Hence if it divides both, then it is $2$. And $2^n$ cannot divide both $2a$ and $2b$ if $n>1$, because then $2^{n-1}>1$ would divide both $a$ and $b$.

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