Question

# If a and b are two consecutive odd positive integers and sum of their squares is 290, then value of a+b is

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Solution

## Let the smaller of two be a, so a+2=b Given condition a2+b2=290⇒a2+(a+2)2=290⇒2a2+4a−286=0⇒a2+2a−143=0⇒a2+13a−11a−143=0⇒a(a+13)−11(a+13)=0⇒(a−11)(a+13)=0⇒a=−13,11∴a=11 (∵a is positive integer)⇒b=11+2=13∴a+b=11+13=24

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