All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Equation in integers (Posted on 2020-12-06) Difficulty: 3 of 5
Determine all pairs of prime numbers (p, q) such that p2 + 5pq + 4q2 is the square of an integer.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 3 of 5 |
p^2+5pq+4q^2 factors into (p+q)*(p+4q).  Any factor in common with p+q and p+4q must also be a factor of their difference 3q.

p and q are coprime then q is coprime to p+q.  For p+q to have a common factor with 3q then p+q must have a common factor with 3.  

Then the greatest common factor of p+q and p+4q is either 1 or 3.

(p+q)*(p+4q) is given to be a perfect square.  Since the common factor is either 1 or 3 then p+q and p+4q are either both perfect squares or both 3 times a perfect square.

In the case the common factor is 1: let p+q=v^2 and p+4q=w^2.  Then 3q=v^2-w^2=(v-w)*(v+w) and 3p=4w^2-v^2=(2w-v)*(2w+v)

Each of v-w and 2w-v is either 1 or 3, making four total cases to check. 
v-w=1 and 2w-v=1 yields q=5/3 and p=7/3; not integers. 
v-w=1 and 2w-v=3 yields q=3 and p=13; solution.   
v-w=3 and 2w-v=1 yields q=11 and p=5; solution.   
v-w=3 and 2w-v=3 yields q=15 and p=21; not primes. 

In the case the common factor is 3: let p+q=3x^2 and p+4q=3y^2.  Then q=y^2-x^2=(y-x)*(y+x) and p=4x^2-y^2=(2x-y)*(2x+y).

Each of y-x and 2x+y must equal 1.  This yields q=5 and p=7, another solution.

  Posted by Brian Smith on 2020-12-06 21:27:42
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 (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information