February 19th, 2016, 10:03 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 02:11 PM. 
February 19th, 2016, 12:57 PM  #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 02:19 PM. 
February 19th, 2016, 04:24 PM  #3 
Member Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 
?

February 19th, 2016, 06:20 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 632 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.


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