perplexus.info :: Just Math : Totient Trial

Totient Trial (Posted on 2016-10-24)

It is observed that Φ(1)=1 and, Φ(2)=1, where Φ(x) denotes Euler's totient function.
So for x=1, it is trivially observed that each of Φ(x) and Φ(x+1) is a perfect square.

(A) What is the next positive integer value of x such that each of Φ(x) and Φ(x+1) is a perfect square?

(B) What is the value of x with 2000 ≤ x ≤ 2100 such that each of Φ(x) and Φ(x+1) is a perfect square?

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

My way

| Comment 1 of 2

The perfect square values: http://oeis.org/A039770/b039770.txt

I just scanned visually for consecutive values then checked with

http://oeis.org/A000010/b000010.txt

Φ(125)=100

Φ(126)=36 so the answer to part A is x=125

Φ(512)=256

Φ(513)=324

Φ(514)=256 so x=512 yields a triple

Φ(1358)=576

Φ(1359)=900

Φ(1728)=576

Φ(1729)=1296

Φ(1970)=784

Φ(1971)=1296

Φ(2047)=1936

Φ(2048)=1024 so the answer to part B is x=2047

Φ(2834)=1296

Φ(2835)=1296 so x=2834 the same perfect square

Posted by Jer on 2016-10-24 13:39:57

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info