See if you can solve Question 3 without using the Intermediate Value Theorem.

See if you can solve Question 3 without using the Intermediate Value Theorem.

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I wonder how many bonus points you would get for fitting #4 in the margin.

Maybe ℵ_0。

Solve Question 3 without using the Intermediate Value Theorem:

Mod 5 @ both sides. The left cannot be 1.

Lies! (For skillfully minimized values of the truth...)

Expression => (x-1)^3 = 2(8-x^2) => x must be odd to have (x-1)^3 even => 8-x^2 is odd, so 2(8-x^2) can not be divisible by 8

Use the formula to find the roots?

haha

number 4

margin

1. gcd(765,114)=gcd(651,114)=gcd(537,114)=gcd(423,114)=gcd(309,114)=gcd(195,114)=gcd(81,114)=gcd(81,33)=gcd(48,33)=gcd(15,33)=gcd(15,18)=gcd(3,15)=gcd(3,12)=gcd(3,9)=gcd(3,6)=gcd(3,3)

The answer is 3.

2. no

3. rational root theorem

4. there isn't enough room

In Q3, factor the LHS as x(x

^{2}-x+3). If x is an integer then x^{2}-x+3 is an odd integer (since x^{2}-x is even) and by completing the square it must be at least 3, so it can't divide 16.The other questions are trivial :)

Fermat's Last Theorem is trivial?

it sure said so somewhere...

For Q3, here's a proof so easy even an undergrad could understand it. x^3 - x^2 + 3x = x(x^2 - x + 3) so for the equation to hold with integer x, x must be a factor of 16, i.e. 1, 2, 4 or 16. A little arithmetic verifies none of these work. QED

You're missing 4 potential factors - the corresponding negatives. they can of course be eliminated by noting that the expressiuon will be negative for negative values of x.

That's what I get for posting math on the internet, there is only an error.

Even easier, x must be +/- 16 since x^2-x+3 is odd.

I have a very elegant proof for Q4, but it does not fit on this margin ;)

Subtract 16 then use the rational root theorem.

Does anyone know how to replicate this comic in LaTeX?

16 = x^3-x^2+3*x = x*(x*(x-1)+3)

Notice that x*(x-1) is even;

Then, x*(x-1)+3 is odd. But 16 has no odd prime factors.

Cool! So therefore you proved that x^3-x^2+3x=16 has no integer roots by using method of contradiction! Q.E.D.

Love the fact Q4 is written in the margin...I wonder how small Andrew Wiles can write?

For 3).. it's same as: (x-1)*(x^2 + 3) = 13..

as RHS is 13, if x is an integer, LHS should be either 13 * 1, or 1 * 13.. Neither is possible with any possible x.

Forgot to mention that it can't be -1*-13 or -13*-1 because (x^2 + 3) >= 3