February 19th, 2016, 09:03 AM  #1 
Member Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 
Find all triple (p, x, y) such that p^x = y^4 + 4 where p is prime number, and x and y are natural numbers. I have only got the factorization by using Sophie Germain's identity: p^x = y^4 + 4 is equivalent to p^x = (y^2  2y + 2)(y^2 + 2y + 2) can't get further with my problem.. Last edited by skipjack; February 19th, 2016 at 01:11 PM. 
February 19th, 2016, 11:57 AM  #2 
Member Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 
Is this proof okay? The difference between the factors on the right side is equal to 4y, and therefore their greatest common divisor also share just 4y. But since p is an odd prime number must also y be odd, and it is none of (yÂ²2y + 2) or (yÂ² + 2y + 2) odd either, so factor the greatest common divisor is odd (and divisor of y). As said y is odd, and thus mutually primisk 2, so that none of the factors is divisible by y either; they give residue 2 by the division of 2. Together, we get that gcd (yÂ²  2y + 2), (yÂ² + 2y + 2) = 1. Consequently, it follows that either (yÂ²  2y + 2) or (yÂ² + 2y + 2) is equal to 1 (they are always positive). This affords, respectively, y = 1 and y = 1; the latter solution is discarded since yâˆˆâ„•. Only solution is (p, x, y) = (5,1,1). Last edited by skipjack; February 19th, 2016 at 01:19 PM. 
February 19th, 2016, 03:24 PM  #3 
Member Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 
?

February 19th, 2016, 05:20 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 638 Thanks: 85 
I thought I learned that the sum of two square numbers can never be factored and have real solutions. For example, if x^2 + 4 = 0, x^2 = 4 and x is imaginary. Based on the problem and how y^4 + 4 was factored, I learned that I was wrong. As for the answer, the first five possible values of y^4 + 4 are 5, 20, 85, 260, and 629. 5 is a solution for p. 20, 85, 260, and 629 each have multiple prime factors, so they can't be solutions. I don't think the problem expects you to look at numbers greater than that.


Tags 
number, theory 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
number theory  gaussrelatz  Math Events  19  November 29th, 2012 01:29 AM 
Number Theory  sxm6719  Computer Science  11  December 19th, 2011 03:45 AM 
Graph theory and number theory  proglote  Number Theory  3  October 30th, 2011 04:20 PM 
number theory  Maria88  Number Theory  2  May 4th, 2010 06:39 AM 
number theory  vinodannu  Number Theory  1  August 17th, 2008 04:20 PM 