number of number pairs

Denote the irrational number $I =( \sqrt n + \sqrt k )^2 = (n+k)+ 2\sqrt{nk}$ An irrational number only exist if and only if $\sqrt{nk}$ is not a perfect square number.

Note that there are $190$ numbers pairs for $1\leq k<n\leq 20$ and counting such all the possible pairs of $(k,n)$ for the number to be irrational is tedious job to do however finding all the possible pairs of $(k,n)$ such the $(\sqrt n + \sqrt k)^2$ is an integer aids us to determine such pairs resulting to irrational number and for which $\sqrt{nk}$ should be perfect square number for $n\neq k$

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Considering case 1 :
Seeking for the perfect square number with Product of distinct factors between 1 to 100 We can take as $\begin{aligned}& 1\leq \sqrt{nk}\leq 10 \\& 1\leq nk \leq 100 \implies 4\leq nk \leq 100\end{aligned}$ We see that for $nk = 4, 9, 16$ and $n\neq k$ , $nk$ yields a perfect square number for the pairs $(k,n) = (1,4),(1,9),(2,8),(1,16)$ and there exist no pairs for $nk =25,49,81$ in the defined range since the one of divisor $n>20$ also they are perfect square number of odd numbers which is only yield if $n=k$ .

Also note that for $nk =36,64,100$ possible pairs $(k,n)= (2,18),(4,9),(3,12),(4,16),(5,20)$ respectively corresponds $nk$ to a perfect square numbers. So numbers of pairs of numbers of $(n,k)$ are $9$ .

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Considering case 2 : Seeking the perfect square number in between 100 to 400 we can directly reject the odd numbers in between 10 to 20. Only for $nk = 144$ pairs of integers are defined for $n<20$ that are $(9,16),(8,18)$ however, for $\sqrt{nk} = 14, 16,18,20$ the divisor of $nk, n>20$ .

$\therefore$ total pairs of $(n,k) = 190 -11 = \boxed{179}$ .