i came across this answer and i saw the given solution but i can not understand how it proves the given problem. Ok i get that $lcm(5,7)= 35$ and it is the same as the $(mod 35)$. Please can someone help me?
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Where are your problems exactly with the the solutions here:http://math.stackexchange.com/questions/755125/how-do-i-prove-that-17n-12n-24n19n-equiv-0-pmod35 ? – Dietrich Burde May 02 '14 at 12:54
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Chinese rem theorem. – rah4927 May 02 '14 at 12:56
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First of all i do not understant if this is the hole solution or just a small part? – user147345 May 02 '14 at 13:07
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1@user147345 Welcome to Math.SE! If you don't understand something in that solution, you should be asking your question as a comment there, rather than as a whole new question. – Nick Peterson May 02 '14 at 13:14
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@Nicholas, I think a user with only 1 point can't leave comments. – Gerry Myerson May 02 '14 at 13:20
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@NicholasR.Peterson i could not leave a comment on that answer.... – user147345 May 02 '14 at 13:24
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Hint $\ \{17,19\}\equiv \{12,24\}\ {\rm mod}\ 5,7\ $ and $\,f(x,y)=x^n\!+y^n$ is symmetric $\,f(x,y)=f(y,x)$
thus $\,f(17,19)\equiv f(12,24)\ {\rm mod}\ 5,7\,$ so $\,5,7\mid f(17,19)\!-\!f(12,24)\,\Rightarrow\,$ so too does $\,{\rm lcm}(5,7)=35$
Bill Dubuque
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