`( a - b\sqrt{2023})/(b-c\sqrt{2023})  in  Q`

Đặt `( a - b\sqrt{2023})/(b-c\sqrt{2023}) = m/n ; ( m ; n ) = 1  ( m in Z ; n \ne 0 in Z )`

`na - nb\sqrt{2023} = mb - mc\sqrt{2023}`

`mc\sqrt{2023} - nb\sqrt{2023} = mb - na`

`\sqrt{2023}( mc - nb ) = mb - na`

`Do  a ; b ; c ; m ; n in Z => mb - na ; mc - nb  in  Z`

`\sqrt{2023} in I => \sqrt{2023}( mc - nb ) in I`

`Dấu = xảy  ra:  {( mc - nb = 0 ),( mb - na = 0 ):}`

`=> m/n = b/c = a/b => ac = b^2  ( 1 )`

Ta có: `a^2 + b^2 + c^2`

`= a^2 + 2b^2 + c^2 - b^2`

`= a^2 + 2ac + c^2 - b^2`

`= ( a - b + c )( a + b + c )`

`Do  ( a - b + c )( a + b + c )` là số nguyên tố 

và  `a ; b ; c in N`* `=> a - b + c < a + b + c`

`=> a - b + c = 1 => a + c - 1 = b  ( 2 )`

`( 1 ) ; ( 2 ) => ac = a^2 + c^2 + 1 + 2ac - 2a - 2c`

`a^2 + c^2 + ac - 2a - 2c + 1 = 0`

`( a - 1 )^2 + c( c + a - 2 ) = 0`

`Do  c ; a in N`* `=> c ; a >= 1 => c( c + a - 2 ) >= 0 ; ( a - 1 )^2 >= 0`

`Dấu = xảy  ra:  a = c = 1 => b = 1`

`Vậy  ( a ; b ; c ) = ( 1 ; 1 ; 1 )`