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

May 2009
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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
 
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Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, the undertaker!

I baby-talked my way through it . . .


Prove that if \(\displaystyle p,q,r, s\) are odd integers,

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

Note that:

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

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


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


Suppose \(\displaystyle x\) is even.

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

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

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



Suppose \(\displaystyle x\) is odd.

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

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

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

 
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Nov 2009
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Wellington
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 \(\displaystyle \mathrm{even}\).

Otherwise, (Clapping)