SOLUTION: Can you please help to answer the following question: . Find all positive integers a and b such that: a^2+b^2-7=ab

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Question 12201: Can you please help to answer the following question:
.
Find all positive integers a and b such that:
a^2+b^2-7=ab

Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
At first,note that a^2+b^2-7=ab -->a^2+b^2 = ab +7.
So, if a,b are positive then a^2+b^2 > 7.

Convert a^2+b^2-7=ab to a%5E2+-2ab%2Bb%5E2+=+7+-+ab i.e.
+%28a-b%29%5E2+=+7+-ab+%3E=+0+. Hence, ab+%3C=+7

Also, notice a & b are symmetric.
[note a,b are positive integers with +a%5E2+%2B+b%5E2+%3E7+ and +0+%3C+ab+%3C=7+]
Consider the table below (using Excel or direct computation)
a b a^2+b^2-ab
1 3 7
1 4 13
1 5 21
1 6 31
1 7 43
2 2 4
2 3 7 [No need to check a >= 3, why ?]
The only pairs (a,b) with +a%3C=+b satisfying a^2+b^2-ab = 7 are
(1,3) and (2,3). Hence,we also have (3,1) and (3,2)
So,there are 4 pairs of positive integer solutions:
(1,3),(2,3), (3,1) and (3,2)

Kenny