Question 1195834
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*** Note that another tutor has pointed out that my solution below overlooks
*** two other solutions, (3,2) and (2,3) -- so there are 6 ordered pairs that
*** satisfy the given inequalities, not just 4.<br>
First, notice that, by symmetry, if (m,n) satisfies all three conditions then (n,m) does also.<br>
Next, since a^2 and b^2 are non-negative, conditions 2 and 3 mean a and b are positive; and condition 1 means a and b are both less than 4.<br>
So we know a and b are both either 1, 2, or 3.  There aren't a lot of ordered pairs to look at, so a "brute force" solution (by simply trying all the ordered pairs with both coordinates 1, 2, or 3) will be easier than a formal algebraic solution.<br>
Trial and error show that (1,1) and (2,2) satisfy the conditions, but (3,3) violates condition 1.<br>
Trial and error shows that (2,1) satisfies the conditions, so (1,2) does also.  And similar trial and error shows that (3,1) and (1,3) do not satisfy all the conditions.<br>
So we have found all the ordered pairs of integers that satisfy all three conditions: (1,1), (2,2), (1,2), and (2,1).<br>
ANSWER: 4<br>
You can easily verify this answer by using x and y instead of a and b and graphing all three equations (not the inequalities) on desmos.com.  The ordered pairs that satisfy all three conditions are the lattice points that are inside all three circles.<br>
Of course, you could also find the answer by doing the graphing first, without the logical analysis I showed in the first part of my response.<br>