Question 1138622
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Plan of attack:<br>
Given the requirement {{{(a+1)/b}}} equal to an integer, choose integer values for a and determine the possible integer values for b; for those possible values of b, determine whether {{{(b+2)/a}}} is also an integer.<br>
If a = 1...<br>
{{{(a+1)/b = 2/b}}}
The possible values for b are 1 and 2.
{{{(b+2)/a = (b+2)/1}}} is an integer for both these values of b.
For b=1, {{{(b+2)/a = 3/1 = 3}}}  -->  (a,b) = (1,1) is a solution
For b=2, {{{(b+2)/a = 4/1 = 4}}}  -->  (a,b) = (1,2) is a solution<br>
If a = 2...<br>
{{{(a+1)/b = 3/b}}}
The possible values for b are 1 and 3.
{{{(b+2)/a = (b+2)/2}}} is not an integer for either b=1 or b=3.<br>
If a = 3...<br>
{{{(a+1)/b = 4/b}}}
The possible values for b are 1, 2, and 4.
{{{(b+2)/a = (b+2)/3}}} is an integer for both b=1 and b=4.
For b=1, {{{(b+2)/a = 3/3 = 1}}}  -->  (a,b) = (3,1) is a solution
For b=4, {{{(b+2)/a = 6/3 = 2}}}  -->  (a,b) = (3,4) is a solution<br>
If a = 4...<br>
{{{(a+1)/b = 5/b}}}
The possible values for b are 1 and 5.
{{{(b+2)/a = (b+2)/4}}} is not an integer for either b=1 or b=5.<br>
If a = 5...<br>
{{{(a+1)/b = 6/b}}}
The possible values for b are 1, 2, 3, and 6.
{{{(b+2)/a = (b+2)/5}}} is an integer for b=3.
For b=3, {{{(b+2)/a = 5/5 = 1}}}  -->  (a,b) = (5,3) is a solution<br>
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I haven't been able to come up with a formal proof; but it is clear to me that there will be no solutions with values of a greater than 5.  So the complete set of ordered pairs of positive integers (a,b) for which both {{{(a+1)/b}}} and {{{(b+2)/a}}} are both integers is
(1,1); (1,2); (3,1); (3,4); (5,3)