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Alone

07/04/2018 at 13:53
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Prove that A=(x+y)(x+2y)(x+3y)(x+4y)+y4 is a square of a number




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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.


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