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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest658? – Krishna Sharma · 2 years, 3 months ago

Log in to reply

– Kandarp Singh · 2 years, 2 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, 3 months ago

Log in to reply

– Krishna Sharma · 2 years, 3 months ago

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