Question

Tìm các cặp số a,b nguyên dương thỏa mãn \(\left(a+3b-2\right)^2=a^2+9b^2+4\)

Answer 1

Ta có:

`(a+3b-2)^2=a^2+9b^2+4`

`a^2+(3b)^2+(-2)^2+2*a*3b+2*a*(-2)+2*(3b)*(-2)=a^2+9b^2+4`

`a^2+9b^2+4+6ab-4a-12b=a^2+9b^2+4`

`6ab-4a-12b=0`

`(6ab-12b)-4a=0`

`6b(a-2)-4a+8=8`

`6b(a-2)-4(a-2)=8`

`3b(a-2)-2(a-2)=4`

`(3b-2)(a-2)=4`

Vì: `a,b\inZ^+` do đó: `3b-2,a-2\in Ư(4)={1;-1;2;-2;4;-4}`

Mà: `3b-2>=-2` do đó:

`TH1:3b-2=1`

`3b=1+2=3`

`b=3/3=1`

Suy ra: `a-2=4`

`->a=4+2=6`

`TH2:3b-2=-1`

`3b=-1+2=1`

`b=1/3(L)`

`TH3:3b-2=-2`

`3b=-2+2=0`

`b=0/3=0`

Suy ra: `a-2=-2`

`a=-2+2=0`

`TH4:3b-2=2`

`3b=2+2=4`

`b=4/3(L)`

`TH5:3b-2=4`

`3b=4+2=6`

`b=6/3=2`

Suy ra: `a-2=1`

`a=1+2=3`

Vậy: `...`