Giải thích các bước giải:

Ta có:

x2+2y2−2xy+4x−3y−26=0x2+2y2−2xy+4x−3y−26=0

→x2−2x(y−2)+(y−2)2+2y2−3y−26−(y−2)2=0→x2−2x(y−2)+(y−2)2+2y2−3y−26−(y−2)2=0

→(x−y+2)2+y2+y−30=0→(x−y+2)2+y2+y−30=0

→4(x−y+2)2+4y2+4y−120=0→4(x−y+2)2+4y2+4y−120=0

→4(x−y+2)2+4y2+4y=120→4(x−y+2)2+4y2+4y=120

→4(x−y+2)2+4y2+4y+1=121→4(x−y+2)2+4y2+4y+1=121

→4(x−y+2)2+(2y+1)2=121→4(x−y+2)2+(2y+1)2=121

→4(x−y+2)2≤121→4(x−y+2)2≤121

→−5≤x−y+2≤5→−5≤x−y+2≤5

→(x−y+2)2≤25→(x−y+2)2≤25

Vì (x−y+2)2(x−y+2)2 là số chính phương

→(x−y+2)2∈{0,1,4,9,16,25}→(x−y+2)2∈{0,1,4,9,16,25}

→(2y+1)2∈{121,117,105,85,57,21}→(2y+1)2∈{121,117,105,85,57,21}

Vì (2y+1)2(2y+1)2 là số nguyên tố

→{(x−y+2)2=0(2y+1)2=121→{(x−y+2)2=0(2y+1)2=121

→{x−y+2=02y+1=±11→{x−y+2=02y+1=±11

→{x=y−2y=5hoặcy=−6→{x=y−2y=5hoặcy=−6

→(x,y)∈{(3,5),(−8,−6)}