Answer of this question

Let the smaller odd positive integers be x
Since the law is consecutive odd
Given

Both integers are smaller than 10,
I.e.   x<10 &  x 2<10
                      x<10-2
                      x<8
Since x<10&x<8
           x<8

Sum of the two integers is more than 11.
therefore, x (x 2)>11
                  2x 2>11
                  2x>11-2
                  2x>9
                  x>9/2
                  x>4.5   
Hence , x>4.5&x<8.

Integers greater than 4.5 but less than 8 are 5,6,7
Odd integers are 5,6,7
Hence x=5 or x=7

x.     x 2      Pair(x 2)
5.    5 2=7     (5,7)
7.  7 2=9       (7,9)  

Thus different set of positive integer possible are (5,7),(7,9).         

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Written on June 23, 2020, 9:30 a.m.