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February 19th, 2016, 10:03 AM   #1
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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.
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February 19th, 2016, 12:57 PM   #2
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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 02:19 PM.
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February 19th, 2016, 04:24 PM   #3
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February 19th, 2016, 06:20 PM   #4
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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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