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March 31, 2015

March 31, 2015

Posted by **ayan** on Wednesday, April 24, 2013 at 12:11am.

- Algebra/Number Theory -
**David**, Wednesday, April 24, 2013 at 7:49pmSolution 1: Since a2>b2+22, thus a>b. Since a and b are consecutive positive integers, we have a=b+1. Substituting in the above expression, we have (b+1)2=b2+2b+1>b2+22⇒b>22−12⇒b≥11. Thus a+b=2b+1≥2×11+1=23.

Solution 2: Since a2>b2+22, thus a>b which implies that a=b+1,a−b=1. Hence a+b=a2−b2a−b>221 ⇒a+b≥23 (since they are integers). We verify that a=11+1=12,b=11 satisfies the inequality, so the minimum possible value of a+b is indeed 23.

- math -
**ayan**, Monday, April 29, 2013 at 1:21am23

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