# No integer solutions to x^10 + px^9 - qx^7 + rx^4 - s = 0

• May 8th 2010, 09:03 PM
the undertaker
No integer solutions to x^10 + px^9 - qx^7 + rx^4 - s = 0
Prove that if p,q,r and s are odd integers, then this equation has no integer roots:

x^10 + px^9 - qx^7 + rx^4 - s = 0
• May 8th 2010, 10:05 PM
Soroban
Hello, the undertaker!

I baby-talked my way through it . . .

Quote:

Prove that if $p,q,r, s$ are odd integers,

then: . $x^{10} + px^9 - qx^7 + rx^4 - s \:=\: 0$ .has no integer roots.

Note that:

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: . $\bigg[(even)^{10} + p(even)^9 + r(even)^4\bigg] - \bigg[q(even)^7 + (odd)\bigg] \;=\;0$

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

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.

• May 8th 2010, 11:59 PM
Bacterius
Geez, that was awesome Soroban. Such a simple technique for such a seemingly intractable problem ! Too bad I can't triple-thank you for this one.

The only improvement I can suggest is putting the "even" and "odd" tags between \mathrm{}, like $\mathrm{even}$.

Otherwise, (Clapping)