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We consider the Euler's totient function $\varphi(n)$ for integers $n\geq 1$. I was playing with the condition $$\varphi(12n)=\varphi(6n+a)\tag{1},$$ for different and simple (integer) values of $a$, when I wondered if one can to find some solution of the equation $$\varphi(12n)=\varphi(6n+5).\tag{2}$$

Question. I don't know why for some integers like $a=1,2,3$ or $a=4$ one can find solutions of $(1)$, but for different values likes than our equation $(2)$ it is difficult to find solutions. Do you have an idea about it? Specially I am asking if we can find some integer $n\geq 1$ of $$\varphi(12n)=\varphi(6n+5).$$ Many thanks.

Computational fact. There aren't solutions of $(2)$ in the interval $1\leq n\leq 10^7.\square$

I think that this problem $(2)$ isn't in the literature, any case if you know it, answer this question as a reference request and I try to find and read the literature to know if there exist some solution of $(2)$

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    $\begingroup$ Perhaps something in $\varphi(12n) = {{\varphi(12)\varphi(n)} \dot {gcd(n, 12)}\over{\varphi(gcd(n, 12))}} $ ; 2, 3 and 4 all divide 12 nicely; $ $\endgroup$ – Ahmed Masud Apr 28 '18 at 20:55
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    $\begingroup$ @user243301 The smallest solution (and the only upto $n=10^8$) is $$80\ 538\ 624$$ (See answer below) $\endgroup$ – Peter Apr 29 '18 at 9:44
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    $\begingroup$ The next solution is $$204\ 160\ 584$$ $\endgroup$ – Peter Apr 29 '18 at 12:32
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    $\begingroup$ And the third is $$359\ 345\ 152$$ $\endgroup$ – Peter Apr 29 '18 at 13:02
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    $\begingroup$ And if you want some bigger solutions (not necessarily consecutive ones), here are a few: $986659840$, $2820644204544$, $5390020614144$, $25538078912512$ (and I'm sure @Peter will notice the pattern I exploited to find them :-) ). $\endgroup$ – Peter Košinár May 3 '18 at 21:54
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There is at least one solution as this PARI/GP-program shows :

? for(n=1,10^8,if(eulerphi(12*n)==eulerphi(6*n+5),print(n)))
80538624
?

Upto $10^8$, there is apparently only one solution.

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    $\begingroup$ +1 for persistence! I didn't bother to test it since the OP had already done it up to $10^7$. $\endgroup$ – lhf Apr 29 '18 at 11:06
  • $\begingroup$ Many thanks for your answer, it's incredible. I am going to wait a week if there is more feedback before I am accepting an answer.. And feel free, when you want, in next week to add the table of known solutions of the equation, many thanks. $\endgroup$ – user243301 Apr 29 '18 at 19:19

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