The Proof is Trivial!

Problem 44***

Let $P(n)$ be a polynomial with coefficients in $\mathbb{Z}$. Suppose that $\deg(P)=p$, where $p$ is a prime number. Suppose also that $P(n)$ is irreducible over $\mathbb{Z}$. Then there exists a prime number $q$ such that $q$ does not divide $P(n)$ for any integer $n$.


Problem 45*

Let $p$ be a prime number, $p\ge 3$. Given that the equation $p^{k}+p^{l}+p^{m}=n^{2}$ has an integer solution, then $p \equiv -1 \pmod 8$.

This feels like a challenge to come up with something. Now if only I knew any physics...

This feels like a challenge to come up with something. Now if only I knew any physics...

That's exactly what I've been thinking, but I can't come up with anything either.

Solution 46:

If we are moving at a speed of 2.5, then the speed can be resolved into two components, $2.5cosx$ and $2.5sinx$, with the cosx being the component travelling across the road. As such, the time take to cross the road is:

$\dfrac{7}{2.5cosx}$

In this time, you will have travelled a distance of $7 tanx$

The time taken for the car to reach this point is:

$\dfrac{40+7tanx}{14}$

We want our time travelled to be less, so we cross the road first:

$\dfrac{7}{2.5cosx} < \dfrac{40+7tanx}{14}$

$98 < 100cosx + 17.5sinx$

$98 < kcos(x-y)$, where k is the square root of 100^2+17.5^2 and y is $arctan \dfrac{17.5}{100}$

$\dfrac{98}{k} < cos(x-y)$

$x < arctan\dfrac{17.5}{100}+arccos \dfrac{98}{\sqrt{100^2+17.5^2}}$

However, this is only half of the question. There is also the possibility the car passes by before we reach halfway:

$\dfrac{3.5}{2.5cosx} > \dfrac{42.5+3.5tanx}{14}$

This means that the time taken for us to reach halfway is greater than the time it takes the car to pass by:

$49 > 106.25cosx+8.75sinx$

$49 > \sqrt{106.25^2+8.75^2}cos(x-arctan \dfrac{8.75}{106.25})$

Which gives us the answer:

$x > arccos \dfrac{49}{\sqrt{106.25^2+8.75^2}} + arctan\dfrac{8.75}{106.25}$

x

This needs a small correction before I can put it in the OP.

I think the question should possibly read $p>3$.

Very good You just need to find the negative values for x as well.

How do you guys latex so fast? That would have taken me an hour and would still be riddled with errors.

I have fast fingers

Edited them in.

Use them to fill in the holes in your working do you?

Looks good to me. What did you think of the question?

Use them to fill in the holes in your working do you?

So many to choose from as well

Looks good to me. What did you think of the question?

Do I really have that many holes in my solutions?

If it does, then I'm going to struggle to find an answer within a reasonable timeframe! Anyway what I posted previously was my current ramblings.