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Math Help - Equation with primes in exponent

  1. #16
    Senior Member abhishekkgp's Avatar
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    Re: Equation with primes in exponent

    godelproof is right. i will put it in a slightly different way.

    It's been established that 5|a^n. Assume that n>1. since 5 is a prime we have 25|a^n \Rightarrow 25|(2^p+3^p).

    LEMMA: If k>0 then 25|(2^k+3^k) iff k \equiv 5,15 \, (\, mod \, 25).
    PROOF: Its easy to see that 2^{20} \equiv 3^{20} \equiv 1 \,(\, mod 25).
    Taking k>0, from the above we have 2^k+3^k can leave at most 20 distinct remainders when divided by 25. we put k=1, \, 2,\, 3, \ldots , \, 20 and using computation find that 2^5+3^5 \equiv 2^{15}+3^{15} \equiv 0 \,(\, mod \, 25). from the blue colored statement above we immediately arrive at the required result.

    from the above lemma we have 25|(2^k+3^k) iff k=20m+5 or k=20m+15, m \in \mathbb{Z} \cup \{0 \}. so the only prime value k can take is 5 and for p=5 it can be computationally verified that 2^5+3^5 \neq a^n, \, n>1.

    we have now proved that 25 \not | (2^p+3^p) if p is prime. so 25 \not | a^n contrary to assumption. QED.
    Last edited by abhishekkgp; June 23rd 2011 at 08:23 AM. Reason: mistake pointed out by godelproof rectified.
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  2. #17
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    godelproof is right. i will put it in a slightly different way.

    It's been established that 5|a^n. Assume that n>1. since 5 is a prime we have 25|a^n \Rightarrow 25|(2^p+3^p).

    LEMMA: If k>0 then 25|(2^k+3^k) iff k \equiv 5,15 \, (\, mod \, 25).
    PROOF: Its easy to see that 2^{20} \equiv 3^{20} \equiv 1 \,(\, mod 25).
    Taking k>0, from the above we have 2^k+3^k can leave at most 20 distinct remainders when divided by 25. we put k=1, \, 2,\, 3, \ldots , \, 20 and using computation find that 2^5+3^5 \equiv 2^{15}+3^{15} \equiv 0 \,(\, mod \, 25). from the blue colored statement above we immediately arrive at the required result.

    from the above lemma we have 25|(2^k+3^k) iff k=25m+5 or k=25m+15, m \in \mathbb{Z} \cup \{0 \}. so the only prime value k can take is 5 and for p=5 it can be computationally verified that 2^5+3^5 \neq a^n, \, n>1.

    we have now proved that 25 \not | (2^p+3^p) if p is prime. so 25 \not | a^n contrary to assumption. QED.
    Thanks abhishekkgp! But how do we "immediately" arrive at the required result?

    and I think k=20m+5 or k=20m+15... not 25. please check it again there.
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  3. #18
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Thanks abhishekkgp! But how do we "immediately" arrive at the required result?

    and I think k=20m+5 or k=20m+15... not 25. please check it again there.
    oops oops !! sorry.. you are right.. i will edit it.
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  4. #19
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Thanks abhishekkgp! But how do we "immediately" arrive at the required result?
    take k=20m+r, \, 0 \leq r < 20. now 2^k+3^k=2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25). so if 2^k+3^k \equiv 0 \, (\,mod \, 25) then 2^r+3^r \equiv 0 \,(\, mod \, 25) which gives r=5, \, 15.
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  5. #20
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define {k}^{*} to be the smallest integer>5 such that 1) LT(2^ {k}^{*} +3^ {k}^{*} )=00 or 25 or 50 or 75; 2) LT(2^ {k}^{*} )=LT(2^5); 3) LT(3^ {k}^{*} )=LT(3^5). Such a {k}^{*} is found by inspecting the first 20 terms of the table to be {k}^{*} =25. And LT(2^ {k}^{*} +3^ {k}^{*} )=75. No problem with this?

    Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT ({2}^{5}{2}^{i})=LT(LT(2^5)LT(2^i))=LT(LT(2^ {k}^{*})LT(2^i))=LT( {2}^{{k}^{*}}{2}^{i})=LT( 2^{{k}^{*}+i}), for any i>0. Similarly LT(3^(5+i))=LT( 3^{{k}^{*}+i}), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^( {k}^{*}+i))+LT(3^( {k}^{*}+i)))=LT(2^( {k}^{*}+i)+3^( {k}^{*}+i))

    From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...
    and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f( {k}^{*}-1)}, and any f(k) \in{00,25,50,75} iff there exists 5 \leqk' \leq {k}^{*}-1 such that f(k')=f(k).

    So what is this k' if f(k') \in{00,25,50,75}? Inspect the table we find k'=5 or k'=15.

    So now you see where k=5+20m and k=15+20m come from.
    --------------------------------------------------------------
    If this doesn't convince you. I happily quit

    ---------------------------------------------------------
    EDIT: red letters.


    Now you've finally produced something that looks like a proof (though I still shall check about the distributivity of that LT function, which right now I've not

    time to do and isn't clear to me at once).

    This is all I was asking before.

    Tonio
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  6. #21
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    Re: Equation with primes in exponent

    Quote Originally Posted by tonio View Post
    Now you've finally produced something that looks like a proof (though I still shall check about the distributivity of that LT function, which right now I've not

    time to do and isn't clear to me at once).

    This is all I was asking before.

    Tonio
    Please feel free to work out every detail you want yourself
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    Taking k>0, from the above we have 2^k+3^k can leave at most 20 distinct remainders when divided by 25
    Quote Originally Posted by abhishekkgp View Post
    2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25)

    Notice that this equality requires 2^k+3^k to have EXACTLY 20 distinct remainders, instead of AT MOST 20 !
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  8. #23
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Notice that this equality requires 2^k+3^k to have EXACTLY 20 distinct remainders, instead of AT MOST 20 !
    i don't get it. what goes wrong if there are less than 20 distinct remainders?? may be i am missing a subtle point...
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  9. #24
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    i don't get it. what goes wrong if there are less than 20 distinct remainders?? may be i am missing a subtle point...
    Let's say there were 18 instead of 20 distinct remainders. Then 2^r \cdot (2^{18})^m+3^r \cdot (3^{18})^m \equiv 2^r+3^r \, (\, mod \, 25).

    Actually, I think from 20 distinct remainders divisible by 25 to concluding 2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25) is not that obvious! That's why I need the LT functions and the tables to make it clear. But perhaps you have better argument than mine! Let's hear it
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  10. #25
    Senior Member abhishekkgp's Avatar
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Let's say there were 18 instead of 20 distinct remainders. Then 2^r \cdot (2^{18})^m+3^r \cdot (3^{18})^m \equiv 2^r+3^r \, (\, mod \, 25).
    how is this true??

    Actually, I think from 20 distinct remainders divisible by 25 to concluding 2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25) is not that obvious! That's why I need the LT functions and the tables to make it clear. But perhaps you have better argument than mine! Let's hear it
    consider the following:
    2^{20} \equiv 1 \,(mod25) \Rightarrow (2^{20})^m \equiv 1  \Rightarrow 2^r \cdot (2^{20})^m \equiv 2^r \,(mod25) .....(1)
    3^{20} \equiv 1 \,(mod25) \Rightarrow (3^{20})^m \equiv 1 \Rightarrow 3^r \cdot (3^{20})^m \equiv 3^r \, (mod25) .....(2)
    so 2^r \cdot (2^{20})^m + 3^r \cdot (3^{20})^m \equiv 2^r+3^r \,(mod25)
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    consider the following:
    2^{20} \equiv 1 \,(mod25) \Rightarrow (2^{20})^m \equiv 1  \Rightarrow 2^r \cdot (2^{20})^m \equiv 2^r \,(mod25) .....(1)
    3^{20} \equiv 1 \,(mod25) \Rightarrow (3^{20})^m \equiv 1 \Rightarrow 3^r \cdot (3^{20})^m \equiv 3^r \, (mod25) .....(2)
    so 2^r \cdot (2^{20})^m + 3^r \cdot (3^{20})^m \equiv 2^r+3^r \,(mod25)
    right. But again, say if 2^k=3^k=1(mod25) for some k=18, then all goes with 18, too. have to check for all k<20, too. that's what i mean by exactly 20!
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  12. #27
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    right. But again, say if 2^k=3^k=1(mod25) for some k=18, then all goes with 18, too. have to check for all k<20, too. that's what i mean by exactly 20!
    yes i had to write a code to find the remainders of 2^k+3^k for k=1,2,3,4,...,20. since 2^k=1 (mod25) is false if k=18 hence it won't work with 18.
    do you think that there is a gap in the proof in post#16??
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    yes i had to write a code to find the remainders of 2^k+3^k for k=1,2,3,4,...,20. since 2^k=1 (mod25) is false if k=18 hence it won't work with 18.
    do you think that there is a gap in the proof in post#16??
    Here's how I see it: You've proved 2^(r+20m)+3^(r+20m)=2^r+3^r(mod25). There's no problem with this. But you did NOT prove 2^(r+km)+3^(r+km) \neq2^r+3^r(mod25) for k<20. So you can NOT conclude the "only if" part of this following statement:

    Quote Originally Posted by abhishekkgp View Post
    we have 25|(2^k+3^k) iff k=20m+5 or k=20m+15, m \in \mathbb{Z} \cup \{0 \}
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  14. #29
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Here's how I see it: You've proved 2^(r+20m)+3^(r+20m)=2^r+3^r(mod25). There's no problem with this. But you did NOT prove 2^(r+km)+3^(r+km) \neq2^r+3^r(mod25) for k<20. So you can NOT conclude the "only if" part of this following statement:
    its hard to explain idea's on internet and it takes a lot of time creating a decent post. i guess you are convinced with your proof( using the LT thingy..) and i am convinced that mine(which is inspired from your method) is correct too but i am not able to explain it here. lets close it. it was a good question.
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    its hard to explain idea's on internet and it takes a lot of time creating a decent post. i guess you are convinced with your proof( using the LT thingy..) and i am convinced that mine(which is inspired from your method) is correct too but i am not able to explain it here. lets close it. it was a good question.
    I agree~

    BTW, Do you believe {2}^{p}+{3}^{p}={a}^{n}\Longrightarrow n=1, for ANY positive integer p? See here http://www.mathhelpforum.com/math-he...ed-183583.html
    Last edited by godelproof; June 25th 2011 at 01:22 AM. Reason: link added
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