Question 12201
 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^2 -2ab+b^2 = 7 - ab}}} i.e.
 {{{ (a-b)^2 = 7 -ab >= 0 }}}. Hence, {{{ab <= 7}}}


 Also, notice a & b are symmetric.
 [note a,b are positive integers with {{{ a^2 + b^2 >7 }}} and {{{ 0 < ab <=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<= 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