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x^10 + px^9 - qx^7 + rx^4 - s = 0 

Prove that if p, q, r, s are odd integers, then the equation has no integer roots

asked Apr 21, 2014 in ALGEBRA 2 by anonymous

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An even integer raised to any positive integral power is even: . \text{(even)}^n \:=\:\text{even}

An odd integer raised to any positive integral power is odd: . \text{(odd)}^n \:=\:\text{odd}


We have:  \bigg[x^{10} + px^9 + rx^4\bigg] - \bigg[qx^7 + s\bigg] \;=\;0


Suppose x is even.

We have:   \bigg[(even)^{10} + p(even)^9 + r(even)^4\bigg] - \bigg[q(even)^7 + (odd)\bigg] \;=\;0   [Since s is odd]

                \underbrace{\bigg[(even) + (even) + (even)\bigg]}_{\text{(even)}} - \underbrace{\bigg[(even) + (odd)\bigg]}_{\text{(odd)}} \;=\;0

And the difference of an even number and an odd number cannot be zero.


Suppose x is odd.

We have:  \bigg[(odd)^{10} + p(odd)^9 + r(odd)^4\bigg] - \bigg[r(odd)^4 + (odd)\bigg] \;=\;0

               \underbrace{\bigg[(odd) + (odd) + (odd)\bigg]}_{\text{(odd)}} - \underbrace{\bigg[(odd) + (odd)\bigg]}_{\text{(even)}} \;=\;0

And the difference of an odd number and an even number cannot be zero.

Therefore it is proved that if p, q, r, s are odd integers, then the equation has no integer roots.

answered Apr 23, 2014 by joly Scholar

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