\(\Leftrightarrow\left(x-3\right)^2+7=y^4\)
\(\Leftrightarrow y^4-\left(x-3\right)^2=7\)
\(\Leftrightarrow\left(y^2-x+3\right)\left(y^2+x-3\right)=7\)
y^2-x+3 | -7 | -1 | 1 | 7 |
y^2+x-3 | -1 | -7 | 7 | 1 |
x | 6 | 0 | 6 | 0 |
y^2 | -4(ktm) | -4(ktm) | 4 | 4 |
y | 2;-2 | 2;-2 |
Vậy pt có các cặp nghiệm nguyên là:
\(\left(x;y\right)=\left(6;2\right);\left(6;-2\right);\left(0;2\right);\left(0;-2\right)\)