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by harsh.champ » Fri Feb 05, 2010 2:57 am
Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

1. 0
2. 1
3. 2
4. 3
5. None of the above
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by thephoenix » Fri Feb 05, 2010 3:52 am
IMO b
x=2 and y=1
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by ajith » Fri Feb 05, 2010 3:56 am
harsh.champ wrote:Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

1. 0
2. 1
3. 2
4. 3
5. None of the above
x^6-y^2 = 127

(x^3+y)(x^3-y) =127

since 127 is a prime number and x and y are positive

x^3 + y = 127

x^3 - y = 1

x^3 =64

y = 63

x=4 y=63 and that is the only pair

B for me
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