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Let $\varphi(n)$ the Euler's totient function. The sequence of even integers $n$ such that $$1+4\varphi(n)$$ is a perfect square starts as $$4, 6, 14, 18, 26, 28, 36, 42, 44, 50\ldots$$

Question. I've curiosity about this question: are there infinitely many even integers $n$ such that $1+4\varphi(n)$ is a perfect square? Many thanks.

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    $\begingroup$ If an odd integer $n$ gives a square, then $2n$ does the same. All because in that case $\phi(2n)=\phi(n)$. Therefore the assumption that $n$ should be even is kinda unnecessary. $\endgroup$ Commented Sep 11, 2017 at 12:15
  • $\begingroup$ Then perfect, many thanks for your claims @JyrkiLahtonen $\endgroup$
    – user243301
    Commented Sep 11, 2017 at 13:10

1 Answer 1

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Yes. Let $p>2$ be a prime. Then $\phi(2p^2)=p(p-1)$ and thus $$ 1+4\phi(2p^2)=1+4p(p-1)=(2p-1)^2. $$

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  • $\begingroup$ Very quick, many thanks then. $\endgroup$
    – user243301
    Commented Sep 11, 2017 at 9:18
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    $\begingroup$ The fact that $(2n+1)^2=4n(n+1)+1$ came in handy many times when I was (35+ years ago) in IMO-training. Just trying to match that pattern. Controlling $\phi(n)$ is easier when $n$ has only a few prime factors. $\endgroup$ Commented Sep 11, 2017 at 9:20
  • $\begingroup$ Many thanks for your tricks. $\endgroup$
    – user243301
    Commented Sep 11, 2017 at 9:27

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