6-degree Polynomials

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, $(n')^{\frac{1}{2}}e^{\frac{(k_1+k_2+k_3)i\pi}{6}}$ is imaginary and is obviously not an integer - a contradiction. Thus, $f_{n'}(x)$ is not of degree 3.

Therefore, $n'$ cannot be positive.

We have now exhausted both cases, and together with the fact that $n' \neq 0$ , we can conclude there do not exist any integer $n'$ such that $x^6+n'$ is reducible and $n'$ cannot be expressed in the form $-a^2$ or $b^3$ . Therefore, the $35$ integers above are the only possible integers $n$ such that $x^6+n$ is reducible and $|n|<500$ . Our answer is thus $35$ .

x^6=(x^2)^3=(x^3)^2 The polynomial p_n(x)=x^6+n can be written as a product of two non-constant polynomials with integer coefficients only if n is a cube or 0 minus the square of an integer number. Let A be the set containing the integers from -500 to 500 that are cubes. Let B be the set containing the integers from -500 to 500 that are 0 minus a square. We are interested on the cardinal of the set A reunited with B. The set A has 15 elements: 0,-1,+1,-2^3,+2^3,-3^3,+3^3,...,-7^3,+7^3 (1+2*7=15). The set B has 22 elements: 0,-1,-2^2,-3^2,...,-22^2 (1+22). The intersection of A with B has 2 elements (0,-1). Using the inclusion-exclusion principle, we have 15+22-2=35 integers which satisfy the conditions of the problem.

we shall use $a^3 - b^3 = (a-b)(a^2+b^2+ab)$

and $a^2 - b^2 = (a+b)(a-b)$

and $a^3 + b^3 = (a+b)(a^2+b^2-ab)$

$0$ is an obvious solution.

for $n \geq 1$ , we can see that all perfect cubes under $500$ work.

there are $7$ such values.

for negative values of $n$ , we count the perfect square values of $|n|$ and its perfect cube values and subtract the perfect sixth powers.

there are $22$ perfect square values from $1$ to $484$ .

and $7$ perfect cube values from $1$ to $343$ .

we overcounted $1$ and $64$ .

so in all there are $1 + 7 + 22 + 7 - 2 = 35$ values of $n$ which satisfy this.

proving that it is always irreducible in the cases not involving perfect squares and perfect cubes is easy. we can assume the given polynomial to be the product of two such lower degree polynomials like two cubics will be a case. a biquadratic and a quadratic will be another case. and the last one will be a five degree one and a linear one. we have to compare the coefficients of the different terms on both the sides and arrive at contradictions..

$p_n(x) = x^6+n$ is factorizable if n can be written as the cube or square of a number. Using the identities for factorization:-

$a^3 \pm b^3 = (a \pm b)(a^2 + b^2 \mp ab)$

$a^2 - b^2 = (a + b)(a - b)$

Case 1 : $n = \pm k^3$

$p_n(x) = x^6 \pm k^3 = (x^2)^3 \pm k^3 = (x^2 \pm k)(x^4 + k^2 \mp x^2k)$

Therefore $p_n(x)$ has been written as a product of two non-constant polynomials.

For integer coefficients such that $|n| < 500$ , $k$ must be a non-negative integer such that $|k^3| < 500$ .

$7^3 = 343$ and $8^3 = 512$

Therefore, $k \in [0,7]$ and so $\pm k^3$ can take $15$ values.

Case 2 : $n = - k^2$

$p_n(x) = x^6 - k^2 = (x^3)^2 - k^2 = (x^3 + k)(x^3 - k)$

Again, $p_n(x)$ has been written as a product of two non-constant polynomials.

For integer coefficients $k$ must be a non-negative integer such that $k^2 < 500$ .

$22^2 = 484$ and $23^2 = 529$

Therefore, $k \in [0,22]$ which gives $23$ values for $- k^2$

Final Answer

From the two cases above there are a total of $15 + 23 = 38$ non-distinct values.

For $n = 0, 1, 64$ we see that trivially $0$ and $1$ are common, also $64 = 4^3 = 8^2$ is common

Thus the total number of distinct integer values of $n$ are $38 - 3 = 35$

Polynomial can be factorized if n is positive and perfect cube(we have 7 such numbers till 500) or is zero or if negative then is a perfect square (we have such 22 numbers till 500) or is perfect cube (we have 7 such numbers till 500). But counting in this manner for negative numbers we counted -1, -64 twice as they are perfect cube as well as perfect square. Hence total choices for n are $7+1+22+7-2=35$

Since the equation given is $x^6$ +n, it is evident that to express it as a product of two polynomials then it has to be either of the following forms:

• $a^3$ + $b^3$
• $a^3$ - $b^3$
• $a^2$ - $b^2$

$x^6$ is already of the form $a^3$ & $a^2$ . So to express n as a form of $b^3$ or $b^2$ , n needs to be a perfect cube or a perfect square. Now all we have to do is calculate the no. of perfect cubes and perfect squares lying in the range of (-500,500). We can see that integers ranging from [-7,7] have their cubes lying in (-500,500). That gives us 15 values of n. Now to calculate perfect squares we can see that integers ranging from [0,22] have their square lying in (-500,500). That gives us 23 values of n. But squares of the no. 0,1 & 8 has already been included in the cubes so these cases need to be eliminated, leaving us with 20 values. Therefore we have now total of 15+20 =35 values.

Solution 1: Suppose $p_n(x)=g(x)\cdot h(x),$ where $g$ and $h$ are not constants. The sum of the degrees of $g$ and $h$ is $6$ , and the product of the leading coefficients is $1.$ Because all coefficients are integers, this means that the leading coefficients of $g$ and $h$ are either both $1$ or both $-1.$ In the latter case, we multiply both $g$ and $h$ by $-1$ so that the leading coefficients are $1$ . Also, we can assume, without loss of generality, that $\deg(g)\geq \deg(h).$ Depending on the degree of $g$ and $h$ , we will consider several cases.

Case 1. $\deg(g)=5,$ $\deg(h)=1$ . Then $h(x)=x-a$ for some integer $a$ , so $a$ is a root of $f_n(x),$ thus $n=-a^6$ for some integer $a$ .

Case 2. $\deg(g)=4,$ $\deg(h)=2$ . If $g(x)=x^4+ax^3+bx^2+cx+d$ and $h(x)=x^2+ex+f,$ then multiplying out and comparing coefficients, we get the following system of equations. $\begin{cases} a+e=0\\ f+ae+b=0\\ af+be+c=0\\ bf+ce+d=0\\ cf+de=0\\ df=n \end{cases}$

From the first equation, $e=-a.$ From the second equation, $f=a^2-b$ . So from the third equation, $a^3-ab-ab+c=0,$ therefore $c=2ab-a^3.$ Plugging this into the fourth equation, we get $d=a^2b+b^2-a^4.$ Then the fifth equation becomes $(2ab-a^3)(a^2-b)-(a^2b+b^2-a^4)a=0,$ which simplifies to $2a^3b-3ab^2=0.$ So $ab(2a^2-3b)=0.$

If $a=0,$ then $e=0$ , $c=0,$ and $f=-b.$ So from the fourth equation, $d=b^2=f^2,$ and from the last equation $n=f^3.$ This is the classical sum of cubes formula: if $n=f^3$ for some integer $f$ , then $x^6+f^3=(x^4-fx^2+f^2)(x^2+f)$ .

If $b=0,$ then $f=a^2$ and $d=-a^4,$ so $n=-a^6.$ In this case, $p_n(x)=(x^4+ax^3-a^3x-a^4)(x^2-ax+a^2).$ Note that we do not get new values of $n,$ compared to the case $a=0.$

If $2a^2-3b=0,$ then $a^2=\frac{3}{2}b,$ so $f=\frac{1}{2}b,$ and $d=\frac{1}{4}b^2.$ This implies $n=\frac{1}{8}b^3=f^3$ . So, again, we get no new values of $n.$

Case 3. $\deg(g)=3,$ $\deg(h)=3$ . If $g(x)=x^3+ax^2+bx+c$ and $h(x)=x^3+dx^2+ex+f,$ then we get the following system of equations. $\begin{cases} a+d=0\\ ad+b+e=0\\ f+c+bd+ae=0\\ af+be+cd=0\\ bf+ce=0\\ cf=n \end{cases}$

From the first two equations, $d=-a$ and $e=a^2-b.$ Plugging this into the remaining equations, we get $\begin{cases} f+c+a(a^2-2b)=0\\ (f-c)a+b(a^2-b)=0\\ bf+c(a^2-b)=0\\ n=cf \end{cases}$

If $a=0,$ we get $f+c=0,$ so $n=-c^2$ . Because these $n$ were already found in Case 1, we may assume that $a\neq 0.$ Then, dividing the second equation by $a,$ we get $f-c=\frac{b^2}{a}-ab.$ Adding this to the first equation, we get $2f=ab-a^3+\frac{b^2}{a}$ . On the other hand, subtracting this from the first equation, we get $2c=3ab-a^3-\frac{b^2}{a}.$ Multiplying the equation $bf+c(a^2-b)=0$ and plugging in the above expressions for $2f$ and $2c,$ we get, after simplification, $3a^3b-3ab^4+\frac{2b^3}{a}-a^5=0.$ Multiplying by $a$ and rewriting, we get $b^3=(a^2-b)^3.$ Because $b$ and $a^2-b$ are real, this means $a^2=2b.$ From this, $f+c=0$ , so $n=-f^2,$ for some integer $f$ .

Putting this all together, $p_n$ can be factored if and only $n=m^3$ or $n=-k^2$ for some integer $m$ and non-negative integer $k$ . Recall that $|n|<500.$ So if $n=m^3$ , then $-7\leq m \leq 7$ and if $n=-k^2,$ then $0\leq k \leq 22.$ Because the intersection of these lists is $\{-4^3, -1^3, 0\},$ the answer is $15+23-3=35.$

Solution 2: We start as in solution 1. Suppose $p_n(x)=g(x)\cdot h(x),$ where $g$ and $h$ are not constants. The sum of the degrees of $g$ and $h$ is $6$ , and the product of the leading coefficients is $1.$ Because all coefficients are integers, this means that the leading coefficients of $g$ and $h$ are either both $1$ or both $-1.$ In the latter case, we multiply both $g$ and $h$ by $-1$ so that the leading coefficients are $1$ . Also, we can assume, without loss of generality, that $\deg(g)\geq \deg(h).$

The polynomial $p_n$ has $6$ complex roots, all with absolute value $\sqrt[6]{|n|}$ . Suppose the degree of $h$ is $k$ , which can be $1,2,$ or $3$ . Then the absolute value of the free term of $h$ is the product of absolute values of $k$ roots, thus it is $|n|^{k/6}.$ If this is an integer, then $|n|$ must be either a perfect square (if $k=3$ ) or a perfect cube. Moreover, if $k=3$ , then $n$ cannot be positive, because every cubic polynomial has a real root, and $x^6+n>0$ for positive $n$ .

So $n$ can be $-7^3,-6^3,...,6^3,7^3$ or $-22^2,-21^2,...,-1^2,0$ . These two sets intersect at $-64,\ -1,$ and $0$ , so the answer is $15+23-3=35.$

suppose, p_n(x)=x^6+n=x^6±y^a.

the expressions "x^6+y^a" & "x^6-y^a" can be factorized only for the following values of "a"

1. x^6+y^6=(x^2+y^2) (x^4-x^2 y^2+y^4)
2. x^6-y^6=(x+y) (x-y) (x^2+x y+y^2) (x^2-x*y+y^2)
3. x^6-y^4=(x^3+y^2)*(x^3-y^2)
4. x^6+y^3=(x^2+y) (x^4-x^2 y+y^2)
5. x^6-y^3=(x^2-y) (x^4+x^2 y+y^2)
6. x^6-y^2=(x^3+y)*(x^3-y)

because | n | < 500 and "y" (±y^a=n) must be an integer, for different values of "a" y^6=0,1,64 y^4=0,1,16,81,256 y^3=0,1,8,27,64,125,216,343 y^2=0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,400,441,484

possible values of "n" =y^6=0,1,64 =-y^6=0,-1,-64 =y^4=0,1,16,81,256 =y^3=0,1,8,27,64,125,216,343 =-y^3=0,-1,-8,-27,-64,-125,-216,-343 =-y^2=0,-1,-4,-9,-16,-25,-36,-49,-64,-81,-100,-121,-144,-169,-196,-225,-256,-289,-324,-361,-400,-441,-484

We can get all possible values of "n" by making an union of these 6 set of numbers (I am using the word "union" because some values have occurred more than one time). That gives 35 distinct values of "n".

When n = k^3 x^6 + k^3 = (x^2)^3 + k^3 (it is of form a^3 + b^3, so can be factorized easily)

When n = -k^3 x^6 - k^3 = (x^2)^3 - k^3 (it is of form a^3 - b^3)

when n = -k^2 x^6 - k^2 = (x^3)^2 - k^2 (it is of form a^2 - b^2)

Factorizations a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a^3 + b^3 = (a - b)(a^2 + ab + b^2) a^2 - b^2 = (a + b)(a - b)

or -n is square

$Let\quad p(x)=f(x)g(x)\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ Since\quad p(x)\quad is\quad of\quad degree\quad 6,\quad the\quad sum\\ of\quad the\quad degrees\quad of\quad f(x)\quad \& \quad g(x)=6\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ The\quad 3\quad possible\quad combos\quad are:\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ 3+3\quad (if\quad n<0\quad and\quad \sqrt { \left| n \right| } =integer,exclude\quad 0\quad for\quad now)\\ 2+4\quad (\sqrt [ 3 ]{ n } =int)\\ 1+6\quad (if\quad n<0\quad and\quad \sqrt [ 6 ]{ \left| n \right| } =integer,exclude\quad 0\quad for\quad now)\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ Listing\quad numbers\quad that\quad are\quad \left| n \right| <500\quad and:\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ -tive\quad Perfect\quad Sq.=\quad -{ (1) }^{ 2 },-{ (2) }^{ 2 }...,-{ (8) }^{ 2 },...,-{ (22) }^{ 2 }\\ Perfect\quad Cube=\quad -{ (7) }^{ 3 },...,-{ (4) }^{ 3 },...-{ (1) }^{ 3 },...,{ (7) }^{ 3 }\\ -tive\quad Perfect\quad { 6 }^{ th }=\quad -{ (1) }^{ 6 },-{ (2) }^{ 6 }\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ Notice\quad that\quad -{ (1) }^{ 6 }=-{ (1) }^{ 3 }=-{ (1) }^{ 2 }\quad and\quad -{ (2) }^{ 6 }=-{ (8) }^{ 2 }=-{ (4) }^{ 3 }\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ Adding\quad them\quad up:\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad .\\ 22+15+2-4\quad [-4\quad because\quad of\quad the\quad repetition]\\ =35\\$

My approach was to consider the complex roots of ${ x }^{ 6 }+n\ =\ 0$ . Once you have it in the form $(x - r_{1})(x-r_{2})(x-r_{3})(x-r_{4})(x-r_{5})(x-r_{6})$ , where each $r_{i}$ is a root, you know that every other factor of $\sqrt [ 6 ]{ n }$ must be the product of some group of those terms.

First, let a = -n. This will save us some minus signs later on.

When a is positive, we get the factors

$\\(x-\sqrt [ 6 ]{ a } ) \\(x+\sqrt [ 6 ]{ a } ) \\(x-(\frac { 1 }{ 2 } \sqrt [ 6 ]{ a } +\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ a } i)) \\(x-(\frac { 1 }{ 2 } \sqrt [ 6 ]{ a } -\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ a} i)) \\(x-(-\frac { 1 }{ 2 } \sqrt [ 6 ]{ a } +\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ a } i)) \\(x-(-\frac { 1 }{ 2 } \sqrt [ 6 ]{ a } -\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ a } i))$

(These are $\sqrt [ 6 ]{ a }$ times the sixth roots of unity)

Since we want to multiply terms that will produce polynomials with integer coefficients, each term with a complex part must be multiplied with its conjugate. This narrows the group down to

$\\(x-\sqrt [ 6 ]{ a } ) \\(x+\sqrt [ 6 ]{ a } ) \\({ x }^{ 2 }-\sqrt [ 6 ]{ a } x+\sqrt [ 3 ]{ a } ) \\({ x }^{ 2 }+\sqrt [ 6 ]{ a } x+\sqrt [ 3 ]{ a} )$

From here, we see that any $a$ that is a perfect sixth will work, but we can do better. Grouping the first two and last two terms produces the difference of cubes identity, while grouping the first and fourth, and the second and third produces the difference of squares (note that every perfect sixth is also a perfect cube or square, so we don't need to consider that anymore). No other grouping produces anything useful, so when a is positive (n is negative), then , it must be a perfect square or perfect cube , and n would be the additive inverse of one.

We do the same thing when n is positive (we are switching back to using n). The factors are

$\\(x-\sqrt [ 6 ]{ n } i) \\(x+\sqrt [ 6 ]{ n} i) \\(x-(\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ n } +\frac { 1 }{ 2 } \sqrt [ 6 ]{ n } i)) \\(x-(\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ n} -\frac { 1 }{ 2 } \sqrt [ 6 ]{ n} i)) \\(x-(-\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ n } +\frac { 1 }{ 2 } \sqrt [ 6 ]{ n } i)) \\(x-(-\frac { \sqrt { 3 } }{ 2 } \sqrt [ 6 ]{ n } -\frac { 1 }{ 2 } \sqrt [ 6 ]{ n} i))$

Grouping conjugates together leaves

$\\({ x }^{ 2 }+\sqrt [ 3 ]{ n } ) \\({ x }^{ 2 }+\sqrt { 3 } \sqrt [ 6 ]{ n } x+\sqrt [ 3 ]{ n } ) \\({ x }^{ 2 }-\sqrt { 3 } \sqrt [ 6 ]{ n } x+\sqrt [ 3 ]{ n } )$

$\sqrt { 3 } \sqrt [ 6 ]{ n }$ is an integer when n is 27 times a perfect sixth. But grouping the last two terms together leaves us with the sum of cubes identity, which is more general. , No other grouping produces anything useful. Thus, when n is positive, it must be a perfect cube.

Putting both of the cases together, there are 22 perfect squares and 7 perfect cubes between 0 and 500, with two overlaps at 1 and 64, leaving 27 solutions for negative n, and another 7 for positive n. Adding on the case where n = 0, we have 27 + 7 + 1 = $\boxed{35}$ possibilities for n.

This is my first solution posted on Brilliant, any suggestions welcome!