If a 2 + b b 2 + c is rational, then a + b + c | a b + b c + c aLet a , b , c ∈ N &gt; 0 . Prove that if a 2 + b b 2 + c is a rational number, then a + b + c | a b + b c + c aHints?My approach was to assume that a 2 + b b 2 + c = p q then p b 2 + p c = q a 2 + q bor 2 ( p b − q a ) = q b − p csquaring: 2 ( p b − q a ) 2 = ( q b − p c ) 2 but that doesn't seems to do much.

Question

If             a              2            +      b              b              2            +      c       is rational, then   a  +  b  +  c      |    a  b  +  b  c  +  c  aLet   a  ,  b  ,  c  ∈            N              &amp;gt;        0            . Prove that if            a              2            +      b              b              2            +      c      is a rational number, then   a  +  b  +  c      |    a  b  +  b  c  +  c  aHints?My approach was to assume that            a              2            +      b              b              2            +      c        =      p    q  then  p  b      2    +  p  c  =  q  a      2    +  q  bor      2    (  p  b  −  q  a  )  =  q  b  −  p  csquaring:  2  (  p  b  −  q  a      )    2    =  (  q  b  −  p  c      )    2  but that doesn&#039;t seems to do much.

Kali Galloway · Accepted Answer

Let             a              2            +      b              b              2            +      c        =  r  ∈      Q  Hence,   (  a  −  r  b  )      2    =  r  c  −  bIf   a  −  r  b  ≠  0 so we get a contradiction.Thus,   a  −  r  b  =  0 and   r  c  −  b  =  0 or   a  =  r  b and   c  =      b    r  Id est,            a      b      +      a      c      +      b      c              a      +      b      +      c        =                    b        2                    (        1        +        r        +                  1          r                )                    b              (        1        +        r        +                  1          r                )              =  band we are done!

Let a,b,c in NN>0. Prove that if (a sqrt(2)+b)/(b sqrt(2)+c) is a rational number, then a+b+c|ab+bc+ca. Hints?

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