Let P(n) be a polynomial with coefficients in Z. Suppose that deg(P)=p, where p is a prime number. Suppose also that P(n) is irreducible over 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≥3. Given that the equation pk+pl+pm=n2 has an integer solution, then p≡−1(mod8).
Since p>2, then all of p^(k,l,m) are odd. Since odd + odd + odd = odd, n^2 must be odd, which means n must be odd. If n is odd it must be congruent to either 1,3,5 or 7 mod 8. 1^2 = 1, 3^2 = 9 , 5^2 =25, 7^2 =49. Hence n^2 is congruent to 1 mod 8.
p^(k,l,m) are all odd so are congruent to 1,3,5 or 7 mod 8. If any of k,l,m are even p^(k,l,m) will be congruent to 1 mod 8, by similar reasoning to above. For odd powers they will be congruent to 1,3,5 or 7 mod8. If p is congruent to 3 mod 8 for example, p to an odd power will also be congruent to 3 mod 8.
The congruencies must 'add up' to 1 mod 8, as n^2 is congruent to 1 mod 8. So:
if p is congruent to 1 mod 8:
1mod8 +1mod8 +1mod8 does not equal 1mod8 so this isn't possible.
p is congruent to 3 mod 8:
3mod8 + 3mod8 + 3mod8 = 9mod8 = 1mod 8 so this is possible.
This is a little Mechanics problem that I come up with (you can use a calculator or you can leave your answers exact) :
A car, 2.5m in length, is travelling along a road at a constant speed of 14ms−1. The car takes up the whole of one half of a road of width 7m. You want to cross the road at a constant speed of 2.5ms−1. You will cross the road at an angle θ degrees to the shortest path to the other side of the road, where theta is positive to the right of this line and negative to the left. The car is approaching you from a distance of 40m to the left of this line, keeping to the opposite side of the road. This car is the only car on the road. Find all possible angles θ such that you can cross the road without being hit.
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:
2.5cosx7
In this time, you will have travelled a distance of 7tanx
The time taken for the car to reach this point is:
1440+7tanx
We want our time travelled to be less, so we cross the road first:
2.5cosx7<1440+7tanx
98<100cosx+17.5sinx
98<kcos(x−y), where k is the square root of 100^2+17.5^2 and y is arctan10017.5
k98<cos(x−y)
x<arctan10017.5+arccos1002+17.5298
Also, we can travel towards the car. In this case, the tan(x) becomes -tan(x), which means that our y changes signs, and so the angle must be less than:
x<−arctan10017.5+arccos1002+17.5298
This angle is measured to the left which means we multiply by -1, and so we have the inequality:
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:
2.5cosx7
In this time, you will have travelled a distance of 7tanx
The time taken for the car to reach this point is:
1440+7tanx
We want our time travelled to be less, so we cross the road first:
2.5cosx7<1440+7tanx
98<100cosx+17.5sinx
98<kcos(x−y), where k is the square root of 100^2+17.5^2 and y is arctan10017.5
k98<cos(x−y)
x<arctan10017.5+arccos1002+17.5298
However, this is only half of the question. There is also the possibility the car passes by before we reach halfway:
2.5cosx3.5>1442.5+3.5tanx
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>106.252+8.752cos(x−arctan106.258.75)
Which gives us the answer:
x>arccos106.252+8.75249+arctan106.258.75
Very good You just need to find the negative values for x as well.
Looks good to me. What did you think of the question?
It's very similar to a STEP question I've seen, but I can't remember which paper. DJ will know obviously. I think it could have been posed as a general question, because numbers are yucky.