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Some Prime Squares Sum Cube (Posted on 2023-05-21) Difficulty: 3 of 5
Find all triplet(s) (p, q, r) of prime numbers, that satisfy this equation:
          p3 = p2 + q2 + r2
Justify your answer with valid reasoning.

See The Solution Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

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solution | Comment 2 of 4 |
p^3 - p^2 is divisible by 3 for any integer >0.

The possible cases are q=r=2, or q=r=3, or one of q,r=2 with the other odd, or both of q,r=odd.  

Only the second case allows a solution:  (p,q,r)=(3,3,3). 

**edit**
Careless and incorrect since if p=2 mod 3, p^3 - p^2 is not divisible by 3.  Ignore. 

Edited on May 21, 2023, 1:18 pm
  Posted by xdog on 2023-05-21 12:22:52

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