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a,b are natural no's .
2013+$$a^2$$=$$b^2$$.
Find the minimum possible value of ab?

Note by Kandarp Singh
2 years, 12 months ago

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658? · 2 years, 12 months ago

yeah but how · 2 years, 10 months ago

hi. See that $$a^2-b^2=(a+b)(a-b)=2013$$. And since $$a$$ and $$b$$ are integers, so have to be $$a+b$$ and $$a-b$$. Then you can work out the factors of $$2013$$ to get $$a$$ and $$b$$, and then it's easy to tell which one will give the minimum product. · 2 years, 12 months ago