Crazy Integers

For $(n^2-2)(n^2-20)<0,$ we need exactly 1 term to be negative and 1 to be positive, so $\large\Rightarrow 2<n^2<20$ That means the permissible values of $n$ are $-4, -3, -2, 2, 3, 4,$ for a total of 6 solutions.

Let ${x}$ = $n^{2}$ to give ( ${x}$ -2)( ${x}$ -20). Drawing the quadratic graph produces the following inequality 2 < ${x}$ < 20. Now take the square values to get $n^{2}$ = 4,9,16. Therefore: ${n}$ = $\pm$ 2, $\pm$ 3 and $\pm$ 4

$(n^2 - 2)(n^2 - 20) < 0,\,$ can be split in two cases:

• $(n^2 - 2) < 0; (n^2 - 20) > 0,\,\Rightarrow\,n_{1,2} =\pm\sqrt 2\approx\pm 1.414;\,n_{3,4}=\pm 2\sqrt 5\approx\pm 4.472;\,$

$\{n1 < n < n2\} \cap \{n < n3 \cup n > n4\}=\varnothing;$

• $(n^2 - 2) > 0; (n^2 - 20) < 0,\,\Rightarrow\\\{n < n1 ∪ n > n2\} \cap \{n3 < n < n4\} = \{n3 < n < n1 \cup n2 < n < n4\}=\\\{- 4 \le n \le - 2 \cup 2 \le n \le 4\}\,\Rightarrow\\n = \{-4,\,-3,\,-2,\, 2,\,3,\,4\}\,\Rightarrow\\N_{sol} = - 2 - ( - 4) + 1 + 4 - 2 + 1 = \boxed 6$

(-2sqrt5,-sqrt2) , (sqrt2,2sqrt5)..... is the solution set... so the integers between these nombers are -2,-3,-4,2,3,4........ therefore there are 6 integers which satisfies n.

It is only negative when n^2 is between 2 and 20. There are 3 square numbers between 2 and 20: 4, 9, and 16. N has two possible values, positive and negative, for each of the square numbers. Therefore there are 6 integers that satisfy the inequality.

-2,=3.=4,2,3,4

( $n^{2}$ -2)( $n^{2}$ -20)<0
Therefore one and only one of $n^{2}$ -2 and $n^{2}$ -20 is negative.
2< $n^{2}$ <20
Possible integer values that satisfy this statement are 2, -2, 3, -3, 4 and -4.
Six integers satisfy this equality.

since 2 < n^2 < 20 implies that n = -4 ,-3,-2, 2,3,4,

(n²-2)(n²-20)<0

                  => 2<n²<20

n={-4,-3,-2,2,3,4} So, the number of integers that satisfies is 6.