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If , where is a prime number and and are the natural numbers. Then find the value of .
The answer is 7.
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p x = y 4 + 4 y 2 + 4 − 4 y 2 = ( y 2 − 2 y + 2 ) ( y 2 + 2 y + 2 ) Now, two factors in the RHS are powers of p and since y 2 − 2 y + 2 < y 2 + 2 y + 2 we must have y 2 − 2 y + 2 ∣ y 2 + 2 y + 2 y 2 − 2 y + 2 ∣ y 2 + 2 y + 2 − ( y 2 − 2 y + 2 ) = 4 y Thus 4 y ≥ y 2 − 2 y + 2 ( y − 3 ) 2 ≤ 7 ∣ y − 3 ∣ ≤ 2 1 ≤ y ≤ 5 Trying these values in turn shows that only possibility is y = 1 , which leads to p = 5 and x = 1 and p + x + y = 7 .