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    (Original post by Mladenov)
    Problem 45*

    Let p be a prime number, p> 3. Given that the equation p^{k}+p^{l}+p^{m}=n^{2} has an integer solution, then p \equiv -1 \pmod 8.
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    Can it be true that p \equiv 3 \pmod 8 or p \equiv -1 \pmod 8? If so, then I have a complete proof.
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    (Original post by Star-girl)
    ....
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    The case p \equiv 3 \pmod8 is not possible. Hint:
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    p^{m-k}+p^{l-k}+1 is a perfect square, assuming k < m,l
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    Problem 48*

    f:\mathbb{R}\to\mathbb{R} and w:\mathbb{R}\to\mathbb{R}. Prove that there exist no f,w such that f\big(w(x)\big)=x^2 and w\big(f(x)\big)=x^3, but that there exist f,w
    satisfying f\big(w(x)\big)=x^2 and w\big(f(x)\big)=x^4
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    Does  wf(x)\leq 0 \  \forall x \leq 0  \  and  \ fw(x) \greq 0 \ \forall x \greq 0  \ mean \  f(x) \leq 0 \ \forall x \leq 0 ?
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    (Original post by Mladenov)
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    The case p \equiv 3 \pmod8 is not possible. Hint:
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    p^{m-k}+p^{l-k}+1 is a perfect square, assuming k < m,l
    Thoughts so far:

    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.

    However if we look back at our original condition:

     p^k + p^l + p^m = n^2 we notice that for p>3 not all three powers can be equal as this would lead to, for example:

     3p^k  which can only be a perfect square when p=3, to make an even power on the 3 in its prime factorization. Any other value of p which is a multiple of 3 would not be both prime and square. Also we have shown above that either all three powers are odd, or two are even and one is odd. So we can say without loss of generality  k < l,m .
     p^{k}(1 + p^{l-k} + p^{m-k}) = n^2

    We know:

     1 + p^{l-k} + p^{m-k} is a perfect square if k is even, and must be odd as all three terms are odd. Hence it is congruent to 1mod8. So if p is congruent to 3mod8, we shown above that all powers must be even. This means l-k and m-k is even. This means that inside the bracket is congruent to 3 mod 8 which is irrelevant.

    If we had p being congruent to -1mod8, then k is even and one of l and m is odd, the other even. This means within the bracket you have congruencies of 1mod8, -1mod8 and 1mod8 which add up to 1mod8.
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    (Original post by metaltron)
    ...

    p is congruent to 3 mod 8:

    3mod8 + 3mod8 + 3mod8 = 9mod8 = 1mod 8 so this is possible.


    ...

    However if we look back at our original condition:

     p^k + p^l + p^m = n^2 we notice that for p>3 not all three powers can be equal as this would lead to, for example:

     3p^k  which can only be a perfect square when p=3, to make an even power on the 3 in its prime factorization. Any other value of p which is a multiple of 3 would not be both prime and square. Also we have shown above that either all three powers are even, or two are even and one is odd. So we can say without loss of generality  k < l,m with k even.

     p^k(1 + p^{l-k} + p^{m-k}) = n^2
    Surely you've shown that either all three powers are odd or two are even and one is odd, since p\equiv 3 \ (mod8) \ and \ p^a \equiv 3 \ (mod 8) \Rightarrow a \ is \ odd, \ \forall a (see bold).

    Apart from this point I have the same proof as you but I've been trying to show that it must actually be the latter case that is true in order to complete the proof...
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    (Original post by Star-girl)
    Surely you've shown that either all three powers are odd or two are even and one is odd, since p\equiv 3 \ (mod8) \ and \ p^a \equiv 3 \ (mod 8) \Rightarrow a \ is \ odd, \ \forall a (see bold).

    Apart from this point I have the same proof as you but I've been trying to show that it must actually be the latter case that is true in order to complete the proof...
    Very true, don't know how that happened
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    (Original post by Star-girl)
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    While evaluating the triple integral in spherical polar coordinates, I got to evaluating the integral in the middle (with respect to theta) and by using the substitution u=a\cos \theta , both limits ended up as a and so the integral is just zero. This meant that I ended up integrating zero with respect to phi to get a non-determinable constant... is that supposed to happen?

    I only taught myself a little vector calculus a few days ago, so I'm still a little hazy on some parts.
    I'm wondering if you're having issues with the question because you haven't multiplied the integral by the Jacobian for the change to spherical polar coordinates.

    So you should end up evaluating

    I =  \displaystyle\ \int_{r=0}^1 \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \cos(|\textbf{a}| r\cos\theta) \ r^2 \sin\theta \ \, \mathrm{d} \phi\,\mathrm{d} \theta\,\mathrm{d}r

    What you should end up with is:

    I = \dfrac{4\pi}{|\textbf{a}|^3} \left(\sin(|\textbf{a}|) - |\textbf{a}|\cos(|\textbf{a}|) \right)

    Hope this helps, if you need any help on the steps in-between let me know!
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    This is as far as I got when I tried it.

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    1+p^{l-k}+p^{m-k}=q^2 \Leftrightarrow p^{l-k}+p^{m-k}=(q+1)(q-1)

    Since p is odd, the right hand side is even and is a multiple of 2^3. Therefore in the case where l \neq m (with equality 4 does not divide the right hand side, so this is impossible) where we can assume without loss of generality that m-l \geq 1, p^{m-l}=-1 \mod 8.
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    (Original post by metaltron)
    ...
     1 + p^{l-k} + p^{m-k} is a perfect square if k is even, and must be odd as all three terms are odd. Hence it is congruent to 1mod8. So if p is congruent to 3mod8, we shown above that all powers must be odd. This means l-k and m-k is even. This means that inside the bracket is congruent to 3 mod 8 which is a contradiction.

    ...
    Sorry if I'm being rather pedantic but the bit in bold is not necessarily true: since all 3 terms are odd, then

     1 + p^{l-k} + p^{m-k} \equiv 1,3,5,7 \ (mod \ 8) i.e. it is not restricted to being 1 \ (mod \ 8).
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    (Original post by Star-girl)
    Sorry if I'm being rather pedantic but the bit in bold is not necessarily true: since all 3 terms are odd, then

     1 + p^{l-k} + p^{m-k} \equiv 1,3,5,7 \ (mod \ 8) it is not restricted to being 1 \ (mod \ 8).
    If its a perfect square it will be 1mod8, I showed this above.
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    (Original post by Noble.)
    I'm wondering if you're having issues with the question because you haven't multiplied the integral by the Jacobian for the change to spherical polar coordinates.

    So you should end up evaluating

    I =  \displaystyle\ \int_{r=0}^1 \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \cos(|\textbf{a}| r\cos\theta) \ r^2 \sin\theta \ \, \mathrm{d} \phi\,\mathrm{d} \theta\,\mathrm{d}r

    What you should end up with is:

    I = \dfrac{4\pi}{|\textbf{a}|^3} \left(\sin(|\textbf{a}|) - |\textbf{a}|\cos(|\textbf{a}|) \right)

    Hope this helps, if you need any help on the steps in-between let me know!
    I had multiplied the integral by the Jacobian, so I must have just made some errors along the way. Thanks - I'll try and see where I went astray.

    Interestingly enough, I evaluated the integral in the opposite direction i.e. dr first then  d\theta and finally  d\phi ...

    EDIT: In hindsight perhaps my direction of evaluation for the integral was not wise - I'd better reverse it as you intended.
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    (Original post by metaltron)
    If its a perfect square it will be 1mod8, I showed this above.
    Using the difference of two squares would have simplified it immensely.
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    (Original post by und)
    Using the difference of two squares would have simplified it immensely.
    How did you show that it is not possible with p being congruent to 3mod8. This would mean the part inside the bracket is not a perfect square.
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    (Original post by metaltron)
    If its a perfect square it will be 1mod8, I showed this above.
    Yes, but my qualm is that you haven't actually shown that it must be a perfect square first...
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    (Original post by Star-girl)
    Yes, but my qualm is that you haven't actually shown that it must be a perfect square first...
    Yeah that's because the proof isn't complete yet.
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    (Original post by metaltron)
    How did you show that it is not possible with p being congruent to 3mod8. This would mean the part inside the bracket is not a perfect square.
    It's clear that the RHS contains an even power of p if n is an integer, and therefore so does the LHS.
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    (Original post by metaltron)
    Yeah that's because the proof isn't complete yet.
    Ah right - my bad. That's what I've been trying to show with little success.
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    If I were to complete my proof it would involve showing by exhaustion that the only a<8: a^b=-1 \mod 8 for some b is a=7.
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    (Original post by und)
    If I were to complete my proof it would involve showing by exhaustion that the only a<8: a^b=-1 \mod 8 for some b is a=7.
    Can't you do that in about one step? 1^2=3^2=5^2=7^2=1mod8, so the only a s.t. a^k=7mod8 for some k is 7.
 
 
 
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