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FC Alan Walker 08/04/2018 at 03:24
We have: \(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^{\text{4}}\)
\(=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4\)
\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)
\(=\left[\left(x^2+5xy+5y^2\right)-y^2\right]\left[\left(x^2+5xy+5y^2\right)+y^2\right]+y^4\)
\(=\left(x^2+5xy+5y^2\right)^2-y^4+y^4\)
\(=\left(x^2+5xy+5y^2\right)^2\)
So A is a square number.
Alone selected this answer. -
¤« 08/04/2018 at 15:05
We have: A=(x+y)(x+2y)(x+3y)(x+4y)+y4
=[(x+y)(x+4y)][(x+2y)(x+3y)]+y4
=(x2+5xy+4y2)(x2+5xy+6y2)+y4
=[(x2+5xy+5y2)−y2][(x2+5xy+5y2)+y2]+y4
=(x2+5xy+5y2)2−y4+y4
=(x2+5xy+5y2)2
So A is a square number.