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    (Original post by Mladenov)
    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.
    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.

    p is congruent to 5 mod8 :

    5mod8 + 5mod8 + 5mod8 = 15mod8 = 7mod 8 not required.

    5mod8 + 5mod8 + 1mod8 = 11mod8 = 3mod8 not required.

    5mod8 + 1mod8 + 1mod8 = 7mod8 not required.

    1mod8 + 1mod8 + 1mod8 = 3mod8 not required, so p can't be congruent to 5mod8.

    For p is congruent to 7mod8:

    7mod8 + 1mod8 + 1mod8 = 9mod8 = 1mod8, hence it is possible.

    So either:

    p is congruent to 3 mod 8 or -1mod8.
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    (Original post by shamika)
    :rofl:

    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.
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    (Original post by shamika)
    :rofl:

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

    (Original post by ukdragon37)
    That's exactly what I've been thinking, but I can't come up with anything either.

    Well all of particle physics is to do with symmetry groups, which slightly related to sets (isn't it?), so....there is potential.

    Heck. The Higgs mechanism is to do with symmetry breaking.
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    Problem 46

    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  \theta 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  \theta such that you can cross the road without being hit.
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    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}}

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

    This angle is measured to the left which means we multiply by -1, and so we have the inequality:

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

    Finally, if we choose to travel towards the car, tan(x) becomes -tan(x), and so we have an angle of:

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

    We multiply by -1, and get the final solution of:

     x< -arccos \dfrac{49}{\sqrt{106.25^2+8.75^2  }} + arctan\dfrac{8.75}{106.25}
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    (Original post by DJMayes)
    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}
    Very good You just need to find the negative values for x as well.
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    (Original post by metaltron)
    x
    I think the question should possibly read p>3.
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    (Original post by DJMayes)
    Solution 43
    This needs a small correction before I can put it in the OP.
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    (Original post by und)
    This needs a small correction before I can put it in the OP.
    It doesn't need putting in; it's essentially identical to your solution anyway so there's no point having both.
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    (Original post by und)
    I think the question should possibly read p>3.
    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.
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    How do you guys latex so fast? That would have taken me an hour and would still be riddled with errors.
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    (Original post by metaltron)
    Very good You just need to find the negative values for x as well.
    Edited them in.
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    (Original post by bananarama2)
    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 :sexface:
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    (Original post by Felix Felicis)
    I have fast fingers :sexface:
    Use them to fill in the holes in your working do you?
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    (Original post by DJMayes)
    Edited them in.
    Looks good to me. What did you think of the question?
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    (Original post by bananarama2)
    Use them to fill in the holes in your working do you?
    I felt the heat from that burn all the way over here!

    (Original post by metaltron)
    Looks good to me. What did you think of the question?
    I think that the idea of the question is a very good one. However, it might be an idea to try find some slightly nicer numbers. :lol:
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    (Original post by bananarama2)
    Use them to fill in the holes in your working do you?
    (Original post by Lord of the Flies)
    So many to choose from as well :sexface:
    Do I really have that many holes in my solutions? Would you mind pointing one or two out?
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    (Original post by metaltron)
    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. :yucky:
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    (Original post by Felix Felicis)
    Do I really have that many holes in my solutions?
    I was just trying to make a joke but it didn't quite work, as often.
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    (Original post by metaltron)
    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.
    I'm not actually sure, it's just what I thought at first and then LotF's example with p=3 sort of cemented that belief.
 
 
 
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