a new solution to Subbarao's congruence (Guy's UPINT problem B37)

[Max Alekseyev]

Dear colleagues,

I'm happy to report the discovery of a composite solution n > 4 to Subbarao's congruence:

phi(n) * tau(n) + 2 == 0  (mod n).

This congruence was known to hold for all prime n and for n = 4. 

Here is a new composite solution:

n = 25643985470955911102 = 2 * 61 * 67 * 6803 * 29027 * 15887233.

It has tau(n) = 32 and satisfies the equality 32 * phi(n) + 2 = 31 * n.

Regards,

Max

References:

[1] M. V. Subbarao. On two congruences for primality. Pacific J. Math. 52 (1974), 261-268.
https://msp.org/pjm/1974/52-1/pjm-v52-n1-p26-p.pdf

[2] R. K. Guy. Unsolved Problems in Number Theory. (3rd ed) Springer New York, NY, 2004.

Problem B37.

[Max Alekseyev]

MINOR CORRECTION:

It has tau(n) = 64 and satisfies the equality 64 * phi(n) + 2 = 31 * n.

Actually, the previous statement was messed up with the one for m = n/2, 

for which we have tau(m) = 32 and which satisfies 32 * eulerphi(m) + 1 = 31 * m.

[Charles Greathouse]

That's very nice Max, how did you find it?

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[Max Alekseyev]

Hi Charles,

I used the same algorithm I used to extend sequences involving Euler's totient, e.g. https://oeis.org/A350777

I'm going to describe it in a paper and make it available soon.

Regards,

Max

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[Robert Gerbicz]

Easy problem, a larger solution:

2920753547351330241632880335=5*17*59*173*487*49363*140038472634953

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[Allan Wechsler]

Do I have this right, that UPINT quoted a conjecture that the only nonprime examples were 1 and 4?

I would have expected to see a comment at A062355 (totient * #divs), or perhaps at A046022 (numbers that are 1, 4, or prime).

-- Allan

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[Max Alekseyev]

Hi Allan,

This is what UPINT says about Subbarao's congruences:

Regards,

Max

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[Max Alekseyev]

Hi Robert,

Yes, the next step would be to create a sequence of such composite solutions 4, ...

I'm not yet sure if mine is the second smallest (and yours is the third one), but I plan to establish that some time soon. 

Regards,

Max

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