Question

cho a, b là các số nguyên chứng minh rằng : A=4a(a+b)(a+1)(a+b+1)+b^2 là một số chính phương

Answer 1

\(A=\left(4a^2+4ab\right)\left(a+b+1-b\right)\left(a+b+1\right)+b^2\)

\(=\left(4a^2+4ab\right)\left[\left(a+b+1\right)^2-b\left(a+b+1\right)\right]+b^2\)

\(=\left(4a^2+4ab\right)\left(a+b+1\right)^2-\left(4a^2+4ab\right).b\left(a+b+1\right)+b^2\)

\(=4a^2\left(a+b+1\right)^2+4ab\left(a+b+1\right)^2-4ab\left(a+b+1\right).\left(a+b\right)+b^2\)

\(=4a^2\left(a+b+1\right)+4ab\left(a+b+1\right)+b^2\)

\(=\left[2a\left(a+b+1\right)+b\right]^2\)