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Totient Trial (Posted on 2016-10-24) Difficulty: 3 of 5
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?

No Solution Yet Submitted by K Sengupta    
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Solution 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
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