The answer is 35.

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For any integer $a$ , we have $p_{(-a^2)}(x)=x^6-a^2=(x^3)^2-a^2=(x^3-a)(x^3+a)$ , so $p_{(-a^2)}$ can be written as a product of two non-constant polynomials with integer coefficients. Similarly, for any integer $b$ , we have $p_{b^2}(x)=x^6+b^3=(x^2)^3+b^3=(x^2+b)(x^4-bx^2+b^2)$ , so $p_{b^3}$ can be written as a product of two non-constant polynomials with integer coefficients. From these, we can generate a list of $n$ such that $p_{n}(x)$ is reducible. For $|n|<500$ , we consider all $n$ of the form $-a^2$ or $b^3$ (where $a$ and $b$ are integers). Since $22^2 < 500 < 23^2$ and $7^3< 500 < 8^3$ , we have the set of all possible $n$ of these forms being $\{0^2,-1^2,...,-22^2\}\cup\{-7^3,-6^3,...,6^3,7^3\}$ . There are $23$ elements in $\{0^2,1^2,...,22^2\}$ and $15$ elements in $\{-7^3,-6^3,...,6^3,7^3\}$ . But the non-positive powers of $6$ are repeated in both sets, and since $2^6<500<3^6$ , the set of repeated elements is $\{0^6,-1^6,-2^6\}$ . Therefore, there are $23+15-3=35$ integers $n$ of the form $-a^2$ or $b^3$ (where $|n|<500$ ), so for these $35$ integers $n$ , $p_n(x)=x^6+n$ is reducible.

Now suppose that there exists an integer $n'$ such that $x^6+n'$ is reducible but $n'$ cannot be expressed in the form $-a^2$ or $b^3$ . Since $p_{n'}(x)$ is reducible, then $p_{n'}(x)=f_{n'}(x)g_{n'}(x)$ for some non-constant polynomials $f_{n'}(x)$ and $g_{n'}(x)$ with integer coefficients. Without loss of generality, let the degree of $f_{n'}(x)$ be not more than the degree of $g_{n'}(x)$ . Since $p_{n'}(x)$ is of degree $6$ , then the sum of the degrees of $f_{n'}(x)$ and $g_{n'}(x)$ is $6$ , and the degree of $f_{n'}(x)$ is $1, 2$ or $3$ . Furthermore, since the leading coefficient of $p_{n'}(x)$ is $1$ , then the leading coefficients of $f_{n'}(x)$ and $g_{n'}(x)$ are either both $1$ or both $-1$ . Without loss of generality, let the leading coefficients of $f_{n'}(x)$ and $g_{n'}(x)$ be both $1$ (otherwise, consider $-f_{n'}(x)$ and $-g_{n'}(x)$ ). It is also clear that the roots of $f_{n'}(x)$ and $g_{n'}(x)$ must be the roots of $p_{n'}(x)$ . Finally, $n'$ cannot be $0$ , so we consider only 2 cases: when $n'$ is negative and when $n'$ is positive.

Case 1: $n'$ is negative

When $n'$ is negative, then the six distinct roots of $p_{n'}(x)$ are $|n'|^{\frac{1}{6}}e^{\frac{ki\pi}{3}}$ , where $k=0, \pm 1, \pm 2, 3$ . Also, we note that, for any positive integer $m$ , if a positive integer $c$ is such that $c^{\frac{1}{m}}$ is an integer $d$ , then $c^{\frac{1}{m}} = d \Rightarrow c = d^m$ , i.e. $c$ can be expressed in the form $d^m$ for an integer $d$ . This is equivalent to saying: for any positive integer $c$ that cannot be expressed in the form $d^m$ (where $d$ is an integer), $c^{\frac{1}{m}}$ is not an integer. Since $n'$ is negative, then it is not of the form $-a^2$ or $b^3$ , and thus $|n'|$ cannot be expressed in the form $a^2$ or $b^3$ . From the above, $|n'|^{\frac{1}{2}}$ and $|n'|^{\frac{1}{3}}$ are therefore not integers. Furthermore, any integer of the form $d^6$ (where $d$ is an integer) can be expressed in the form $b^3$ (by setting $b=d^2$ ), so $|n'|$ cannot be expressed in the form $d^6$ (where $d$ is an integer) and $|n'|^{\frac{1}{6}}$ is not an integer. We now consider three sub-cases, when $f_{n'}(x)$ is of degrees $1,2$ or $3$ .

Case 1A: $f_{n'}(x)$ is of degree 1

If $f_{n'}(x)$ is of degree 1, then $f_{n'}(x)=x-|n'|^{\frac{1}{6}}e^{\frac{k_1i\pi}{3}}$ for some $k_1=0, \pm 1, \pm 2$ or $3$ . Then $|n'|^{\frac{1}{6}}e^{\frac{k_1i\pi}{3}}$ must be an integer, so $e^{\frac{k_1i\pi}{3}}$ is real and equal to $1$ or $-1$ (since $|e^{\frac{k_1i\pi}{3}}|=1$ ). But $|n'|^{\frac{1}{6}}$ is not an integer, so $|n'|^{\frac{1}{6}}e^{\frac{k_1i\pi}{3}}$ cannot be an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 1.

Case 1B: $f_{n'}(x)$ is of degree 2

If $f_{n'}(x)$ is of degree 2, then $f_{n'}(x)=(x-|n'|^{\frac{1}{6}}e^{\frac{k_1i\pi}{3}})(x-|n'|^{\frac{1}{6}}e^{\frac{k_2i\pi}{3}})$ for some distinct $k_1,k_2=0, \pm 1, \pm 2$ or $3$ . The constant term of $f_{n'}(x)$ is then $|n'|^{\frac{1}{3}}e^{\frac{(k_1+k_2)i\pi}{3}}$ . Again, this is an integer, so $e^{\frac{(k_1+k_2)i\pi}{3}}$ is real and equal to $1$ or $-1$ . But $|n'|^{\frac{1}{3}}$ is not an integer, so $|n'|^{\frac{1}{3}}e^{\frac{(k_1+k_2)i\pi}{3}}$ cannot be an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 2.

Case 1C: $f_{n'}(x)$ is of degree 3

If $f_{n'}(x)$ is of degree 3, then $f_{n'}(x)=(x-|n'|^{\frac{1}{6}}e^{\frac{k_1i\pi}{3}})(x-|n'|^{\frac{1}{6}}e^{\frac{k_2i\pi}{3}})(x-|n'|^{\frac{1}{6}}e^{\frac{k_3i\pi}{3}})$ for some distinct $k_1,k_2,k_3=0, \pm 1, \pm 2$ or $3$ . The constant term of $f_{n'}(x)$ is then $|n'|^{\frac{1}{2}}e^{\frac{(k_1+k_2+k_3)i\pi}{3}}$ . Again, this is an integer, so $e^{\frac{(k_1+k_2+k_3)i\pi}{3}}$ is real and equal to $1$ or $-1$ . But $|n'|^{\frac{1}{2}}$ is not an integer, so $|n'|^{\frac{1}{2}}e^{\frac{(k_1+k_2+k_3)i\pi}{3}}$ cannot be an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 3.

Therefore, $n'$ cannot be negative.

Case 2: $n'$ is positive

When $n'$ is positive, then the six distinct roots of $p_{n'}(x)$ are $(n')^{\frac{1}{6}}e^{\frac{ki\pi}{6}}$ , where $k=\pm 1, \pm 3, \pm 5$ . Also, as proven above, for any positive integer $c$ that cannot be expressed in the form $d^m$ (where $d$ is an integer), $c^{\frac{1}{m}}$ is not an integer. Since $n'$ is positive and not of the form $b^3$ , then from the above, $(n')^{\frac{1}{3}}$ is not an integer. Furthermore, like in case 1, $n'$ cannot be expressed in the form $d^6$ (where $d$ is an integer) since it cannot be expressed in the form $b^3$ , so $(n')^{\frac{1}{6}}$ is not an integer. We now consider three sub-cases, when $f_{n'}(x)$ is of degrees $1,2$ or $3$ .

Case 2A: $f_{n'}(x)$ is of degree 1

This case is similar to case 1A. If $f_{n'}(x)$ is of degree 1, then $f_{n'}(x)=x-(n')^{\frac{1}{6}}e^{\frac{k_1i\pi}{6}}$ for some $k_1=\pm 1, \pm 3$ or $\pm 5$ . Then $(n')^{\frac{1}{6}}e^{\frac{k_1i\pi}{6}}$ must be an integer, so $e^{\frac{k_1i\pi}{6}}$ is real and equal to $1$ or $-1$ . But $(n')^{\frac{1}{6}}$ is not an integer, so $(n')^{\frac{1}{6}}e^{\frac{k_1i\pi}{6}}$ cannot be an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 1.

Case 2B: $f_{n'}(x)$ is of degree 2

This case is similar to case 2A. If $f_{n'}(x)$ is of degree 2, then $f_{n'}(x)=(x-(n')^{\frac{1}{6}}e^{\frac{k_1i\pi}{6}})(x-(n')^{\frac{1}{6}}e^{\frac{k_2i\pi}{6}})$ for some distinct $k_1,k_2=\pm 1, \pm 3$ or $\pm 5$ . The constant term of $f_{n'}(x)$ is then $(n')^{\frac{1}{3}}e^{\frac{(k_1+k_2)i\pi}{6}}$ . Again, this is an integer, so $e^{\frac{(k_1+k_2)i\pi}{6}}$ is real and equal to $1$ or $-1$ . But $(n')^{\frac{1}{3}}$ is not an integer, so $(n')^{\frac{1}{3}}e^{\frac{(k_1+k_2)i\pi}{6}}$ cannot be an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 2.

Case 2C: $f_{n'}(x)$ is of degree 3

If $f_{n'}(x)$ is of degree 3, then $f_{n'}(x)=(x-(n')^{\frac{1}{6}}e^{\frac{k_1i\pi}{6}})(x-(n')^{\frac{1}{6}}e^{\frac{k_2i\pi}{6}})(x-(n')^{\frac{1}{6}}e^{\frac{k_3i\pi}{6}})$ for some distinct $k_1,k_2,k_3=\pm 1, \pm 3$ or $\pm 5$ . The constant term of $f_{n'}(x)$ is then $(n')^{\frac{1}{2}}e^{\frac{(k_1+k_2+k_3)i\pi}{6}}$ . But as $k_1,k_2$ and $k_3$ are odd, then $k_1+k_2+k_3$ is odd and $\frac{k_1+k_2+k_3}{6}$ cannot be an integer. This means that $e^{\frac{(k_1+k_2+k_3)i\pi}{6}}$ must not be real, since its imaginary component is $\sin (\frac{k_1+k_2+k_3}{6}\pi )$ and this is $0$ if and only if $\frac{k_1+k_2+k_3}{6}$ is an integer. Therefore, $(n')^{\frac{1}{2}}e^{\frac{(k_1+k_2+k_3)i\pi}{6}}$