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 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 (y²-2y + 2) or (y² + 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 (y² - 2y + 2), (y² + 2y + 2) = 1. Consequently, it follows that either (y² - 2y + 2) or (y² + 2y + 2) is equal to 1 (they are always positive). This affords, respectively, y = 1 and y = -1; the latter solution is discarded since y∈ℕ. 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: 660 Thanks: 87 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. Thanks from Inflekx12 Tags number, theory Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gaussrelatz Math Events 19 November 29th, 2012 01:29 AM sxm6719 Computer Science 11 December 19th, 2011 03:45 AM proglote Number Theory 3 October 30th, 2011 04:20 PM Maria88 Number Theory 2 May 4th, 2010 06:39 AM vinodannu Number Theory 1 August 17th, 2008 04:20 PM

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