Q13/050) Given m,n integers satisfying : √(m+2007)+√(m−325) = n find Max(n) and min (n)

we have

(m+ 2007) – (m- 325) = 2332 ..1

√(m+2007)+√(m−325) = n … (2) (given)

Dividing (1) by (2) we get

√(m+2007)-√(m−325) = 2332/n … (3)

Add (2) and (3) to get 2√(m+2007)= (2332/n +n)

Now RHS has to be even so n is even factor of 2332/2 or 1166

Hence n has to be largest even factor of 1166  or 1166

Now to find the lowest n

put m- 325 = p (which is >=0)

we get √(p+2332)+√p which is >= sqrt(2332) >= 49

is has to be even smallest even factor of 2232 but not < 49

now 2332 = 2 * 2 * 11 *53 hence minimum n = 2 * 53 = 106