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| A Second Degree Puzzle (Posted on 2007-06-30) |
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Show that there does not exist any positive integer pair (a, b) satisfying the equation a2 + 3ab – 2b2 = 156
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Submitted by K Sengupta
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Rating: 4.0000 (1 votes)
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Solution:
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(Hide)
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4a^2 + 12ab – 8b^2 = 624
Or, P^2 – 17b^2 = 624, where P = 2a+3b
Or, P^2 Mod 17 = 12
This is a contradiction since 12 is never a quadratic residue of 17.
Consequently, no positive integer pair (a, b) exists which satisfy the given equation.
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