Number theory problem, fractions and gcd The problem says:If a and b are positive integers such that a + 1 b + b + 1 a is an integer, then show that a + b ≥ gcd ( a , b )Adding 2 a b a b to a + 1 b + b + 1 a yields that ab divides (a+b+1)(a+b), but I haven´t been able to continue from there.

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Number theory problem, fractions and gcd The problem says:If   a and   b are positive integers such that                     a        +        1            b        +                    b        +        1            a       is an integer, then show that       a    +    b    ≥  gcd  (  a  ,  b  )Adding             2      a      b              a      b       to             a      +      1        b    +            b      +      1        a   yields that ab divides (a+b+1)(a+b), but I haven´t been able to continue from there.

Carmelo Payne · Accepted Answer

If   gcd  (  a  ,  b  )  =  1, then clearly       a    +    b    ≥  gcd  (  a  ,  b  ). Now assume   gcd  (  a  ,  b  )  =  d  &amp;gt;  1. Then if             a      +      1        b    +            b      +      1        a    =  n, then by your simplification, we have  (  a  +  b  +  1  )  (  a  +  b  )  =  (  n  +  2  )  a  bNote that       d    2   divides the righthand side, so it must also divide the lefthand side. Since   d divides   a and   d divides   b,   d must divide   a  +  b. Since   a  +  b  +  1 and   a  +  b are coprime, and since   d divides   a  +  b, we must also have       d    2   dividing   a  +  b. In particular,   a  +  b  ≥      d    2      ⟹        a    +    b    ≥  d

Number theory problem, fractions and gcd The problem says: If a and b are positive

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