Solution 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.
a and b are consecutive, positive integers such that a2−b2>22. What is the minimum possible value of a+b?
2 answers
23