a,b are natural no's .

2013+\(a^2\)=\(b^2\).

Find the minimum possible value of ab?

a,b are natural no's .

2013+\(a^2\)=\(b^2\).

Find the minimum possible value of ab?

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TopNewest658? – Krishna Sharma · 2 years, 8 months ago

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– Kandarp Singh · 2 years, 6 months ago

yeah but howLog in to reply

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. – Satvik Golechha · 2 years, 8 months ago

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– Krishna Sharma · 2 years, 8 months ago

In general I simply find max{|a-b|} for ab to be minimumLog in to reply