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Start with a perfect square, denoted as a positive integer n. The root of this square is k, another positive integer. Thus n = k^2 Let t = totient(n)

Is there a way to prove there is no such number? I know that Fermat Primes have totient values that are squares, but what about any perfect squares who have a perfect square as their totient value?

And more generally, expanding this concept with an additional restriction, can we say that a totient value (derived from a square number) will not be part of a Pythagorean Triple?

n = t + a

Where a is another integer, and is the difference between n and its totient value.

a = n - t

In other words, can n, t, and a all be perfect squares?

Furthermore, if this is disprovable, then can we also disprove the case of Pythagorean Quadruples? Where perfect squares must sum together to equal the totient value of a perfect square? Does removing the restriction of it being a Pythagorean Triple/Quadruple, and allowing a to be non-square make this possible?

I am not a mathematician. I'm an Electrical Engineer, and have been exploring the world of Number Theory. I find it fascinating and have posed this question for myself, as I have not seen an answer. I believe it's possible I've missed something simple here and there is an "easy" way to show that such things cannot be. I have my computer checking by brute force if any totients of n are also square, and got up to the first 10,000 perfect squares as having no totient number that is a perfect square.

Any insight is appreciated, and I am sorry for the lack of formatting and formal means of formulating my question. I hope this question is clear. I ultimately want to know if there is any relationship between the Pythagorean Triples/Quadruples(or higher sets) to Euler's Totient Function. Wondering if by "coincidence" there is such a number, or if it is impossible (and why). Thank you for taking the time to think about this with me. This is a related question, but is not my question: Pythagorean triples that "survive" Euler's totient function

user2664280
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1 Answers1

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Besides $1$ there is no such example. If $p$ is the largest prime dividing $k^2$ with exponent $2m$ then $\varphi(k^2)$ is divided by $p$ with exponent $2m-1$.

It's easy to see from a direct calculation.

Oscar Lanzi
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Jakobian
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  • Excellent! Thank you so much. I believe I understand. – user2664280 Jan 02 '21 at 04:08
  • Now I must assess if it is possible for a Totient value (derived from a perfect square) to be expressed as a sum of unique perfect squares. We've ruled out that it cannot be the sum of one square, but can it be the sum of 2 unique perfect squares or 3, and so on? I feel this question is harder to solve, but very well might also be "easy" to prove/disprove. I will be searching for examples, hopefully it will be a quick discovery. Thanks for your input. – user2664280 Jan 02 '21 at 04:42
  • Yes found one example: k = 10 We have: totient(10^2)= 40 = 2^2 + 6^2 – user2664280 Jan 02 '21 at 04:57
  • Now to finalize this and put this entire question to rest, I need to find an example (or disprove) that the remainder a = (n - t) is also the sum of squares (or is itself a square). If so, we would have a Pythagorean Quadruple (or higher) derived from the numbers n, t, and a. I believe this should be possible and will look for an example. – user2664280 Jan 02 '21 at 05:46
  • @user2664280 How do you get $40$ from $10^2$? Perhaps totient is not what I understand it to be. – poetasis Jan 02 '21 at 05:48
  • @poetasis - this is simple; do, for instance in Pari/GP, eulerphi(100) and you get $40$. But perhaps... ;-) – Gottfried Helms Jan 02 '21 at 06:26
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    But from another prime, $q-1$ may give more powers of $p$. Still, that won't happen to the largest prime factor $p$ – Empy2 Jan 02 '21 at 13:27
  • @Empy2 That's true. I wrote "largest prime" at first but second guessed myself and edited it – Jakobian Jan 03 '21 at 00:51
  • I changed the claim to the largest prime factor, which is true and sufficient for the proof. Note that such a factor does not exist for $1$, the sole exception. – Oscar Lanzi Aug 31 '23 at 00:37