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Thread: 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 $\displaystyle 5|a^n$. Assume that $\displaystyle n>1$. since $\displaystyle 5$ is a prime we have $\displaystyle 25|a^n \Rightarrow 25|(2^p+3^p)$.

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

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

    we have now proved that $\displaystyle 25 \not | (2^p+3^p)$ if $\displaystyle p$ is prime. so $\displaystyle 25 \not | a^n$ contrary to assumption. QED.
    Last edited by abhishekkgp; Jun 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 $\displaystyle 5|a^n$. Assume that $\displaystyle n>1$. since $\displaystyle 5$ is a prime we have $\displaystyle 25|a^n \Rightarrow 25|(2^p+3^p)$.

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

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

    we have now proved that $\displaystyle 25 \not | (2^p+3^p)$ if $\displaystyle p$ is prime. so $\displaystyle 25 \not | a^n$ 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 $\displaystyle k=20m+r, \, 0 \leq r < 20$. now $\displaystyle 2^k+3^k=2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25)$. so if $\displaystyle 2^k+3^k \equiv 0 \, (\,mod \, 25)$ then $\displaystyle 2^r+3^r \equiv 0 \,(\, mod \, 25)$ which gives $\displaystyle r=5, \, 15$.
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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 $\displaystyle {k}^{*}$ to be the smallest integer>5 such that 1) LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {k}^{*} $)=00 or 25 or 50 or 75; 2) LT(2^$\displaystyle {k}^{*} $)=LT(2^5); 3) LT(3^$\displaystyle {k}^{*} $)=LT(3^5). Such a $\displaystyle {k}^{*} $ is found by inspecting the first 20 terms of the table to be $\displaystyle {k}^{*} =25$. And LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {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$\displaystyle ({2}^{5}{2}^{i})$=LT(LT(2^5)LT(2^i))=LT(LT(2^$\displaystyle {k}^{*}$)LT(2^i))=LT($\displaystyle {2}^{{k}^{*}}{2}^{i}$)=LT($\displaystyle 2^{{k}^{*}+i}$), for any i>0. Similarly LT(3^(5+i))=LT($\displaystyle 3^{{k}^{*}+i}$), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^($\displaystyle {k}^{*}+i$))+LT(3^($\displaystyle {k}^{*}+i$)))=LT(2^($\displaystyle {k}^{*}$+i)+3^($\displaystyle {k}^{*}$+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($\displaystyle {k}^{*}-1$)}, and any f(k)$\displaystyle \in${00,25,50,75} iff there exists 5$\displaystyle \leq$k'$\displaystyle \leq$$\displaystyle {k}^{*}-1$ such that f(k')=f(k).

    So what is this k' if f(k')$\displaystyle \in${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 $\displaystyle k>0$, from the above we have $\displaystyle 2^k+3^k$ can leave at most $\displaystyle 20$ distinct remainders when divided by $\displaystyle 25$
    Quote Originally Posted by abhishekkgp View Post
    $\displaystyle 2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25)$

    Notice that this equality requires $\displaystyle 2^k+3^k$ 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 $\displaystyle 2^k+3^k$ 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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    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 $\displaystyle 2^r \cdot (2^{18})^m+3^r \cdot (3^{18})^m \equiv 2^r+3^r \, (\, mod \, 25)$.

    Actually, I think from 20 distinct remainders divisible by 25 to concluding $\displaystyle 2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25)$ 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 $\displaystyle 2^r \cdot (2^{18})^m+3^r \cdot (3^{18})^m \equiv 2^r+3^r \, (\, mod \, 25)$.
    how is this true??

    Actually, I think from 20 distinct remainders divisible by 25 to concluding $\displaystyle 2^r \cdot (2^{20})^m+3^r \cdot (3^{20})^m \equiv 2^r+3^r \, (\, mod \, 25)$ 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:
    $\displaystyle 2^{20} \equiv 1 \,(mod25) \Rightarrow (2^{20})^m \equiv 1 \Rightarrow 2^r \cdot (2^{20})^m \equiv 2^r \,(mod25)$ .....(1)
    $\displaystyle 3^{20} \equiv 1 \,(mod25) \Rightarrow (3^{20})^m \equiv 1 \Rightarrow 3^r \cdot (3^{20})^m \equiv 3^r \, (mod25)$ .....(2)
    so $\displaystyle 2^r \cdot (2^{20})^m + 3^r \cdot (3^{20})^m \equiv 2^r+3^r \,(mod25)$
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    Re: Equation with primes in exponent

    Quote Originally Posted by abhishekkgp View Post
    consider the following:
    $\displaystyle 2^{20} \equiv 1 \,(mod25) \Rightarrow (2^{20})^m \equiv 1 \Rightarrow 2^r \cdot (2^{20})^m \equiv 2^r \,(mod25)$ .....(1)
    $\displaystyle 3^{20} \equiv 1 \,(mod25) \Rightarrow (3^{20})^m \equiv 1 \Rightarrow 3^r \cdot (3^{20})^m \equiv 3^r \, (mod25)$ .....(2)
    so $\displaystyle 2^r \cdot (2^{20})^m + 3^r \cdot (3^{20})^m \equiv 2^r+3^r \,(mod25)$
    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)$\displaystyle \neq$2^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 $\displaystyle 25|(2^k+3^k)$ iff $\displaystyle k=20m+5$ or $\displaystyle k=20m+15$, $\displaystyle m \in \mathbb{Z} \cup \{0 \}$
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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)$\displaystyle \neq$2^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 $\displaystyle {2}^{p}+{3}^{p}={a}^{n}\Longrightarrow n=1$, for ANY positive integer p? See here http://www.mathhelpforum.com/math-he...ed-183583.html
    Last edited by godelproof; Jun 25th 2011 at 01:22 AM. Reason: link added
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