Integer factorization

It is obvious that $f_n(x)$ must be either in the form of $(x+a)(x^3+bx^2+cx+d) or (x^2+ax+b)(x^2+cx+d)$ ,

with $a, b, c$ and $d$ are integers.

first case

$f_n(x)=(x+a)(x^3+bx^2+cx+d)$ $\iff f_n(x)=x^4+(a+b)x^3+(ab+c)x^2+(ac+d)x+ad$ then, $a+b=ab+c=ac+d=0$

We get, $a=-b$ , $c=a^2$ , $d=-a^3$ , $n=ad>0$

$n=ad \iff n=-a^4 \leq 0$

which is contradiction to the problem.

second case

$f_n(x)= (x^2+ax+b)(x^2+cx+d)$ $\iff f_n(x)=x^4+(a+c)x^3+(b+d+ac)x^2+(bc+ad)x+bd$ then, $a+c=b+d+ac=bc+ad=0$

We get, $a=-c$ , $n=bd>0$ , $b+d=a^2 \iff a \neq 0$ , $bc+ad=c(b-d)=0 \iff b=d$

$2b=a^2$

We get, $b=2, 8, 18, 32, \ldots$

$n=b^2<1000$

So $n=4, 64, 324$

are the only solutions. So, the result occurs.

We set $n=4y^4$ . By Sophie-Germain identity, we have:

$f_n (x) = x^4+n = x^4+4y^4 = \left(x^2+2xy+2y^2\right)\left(x^2-2xy+2y^2\right)$

And it factorizes as two non-constant polynomials with integer coefficients. Now note that:

$4\cdot 3^4 = 324 < 1000 < 1024 =4\cdot 4^4$

Hence the answer is simply:

$\sum_{i=1}^{3} 4\cdot i^4 = \fbox{392}$

The polynomial can't have a real root, so both factors would have to be quadratic. WLOG they're monic. After looking at the coefficients of $x^3$ and $x$ , it has to be $x^4+n = (x^2-ax+b)(x^2+ax+b).$ So $b^2 = n$ and $a^2 = 2b$ , so $a$ and $b$ are both even and $n = 4(a/2)^4.$ Since $n < 1000,$ there are only three solutions, namely $4 \cdot 1^4, 4 \cdot 2^4, 4 \cdot 3^4.$ The sum of these is $392$ .

x^4+4y^4 is reducible.

Hence n = 4y^4 <1000. y= 1,2,3 n = 4, 64, 324

In order to find $n$ , we need to find its non-constant factors with integer coefficients. Since $x^4+n$ , where $n$ is a positive integer, has no middle term we can assume that the middle term with variable $x^2$ has been cancelled. By looking for two polynomials which will yield to the expression $x^4+n$ , we get $(x^2+2a+2a^2)$ $(x^2-2a+2a^2)$ . Simplifying the expression we get: $x^4+4a^4$ . Therefore $n=4a^4$ where $a$ is any positive integer.

When $a=1$ , $n=4$ .

When $a=2$ , $n=64$ .

When $a=3$ , $n=324$ .

When $a=4$ , $n=1024$ . Since $n<1000$ , 1024 is not a solution and as $a$ gets bigger so as the value of $n$ , therefore, we will only consider $a=1,2,3$ .

Now we have the values of $n$ which are 4, 64, 324.

$4+64+324=392$

If $f_n (x)=x^4 +n$ is composed of two non-constant polynomials with integer coefficients, this means that it is composed of two Quadratic Equation.

Lets say the two equation is $g(x)=x^2+ax+b$ and $h(x)=x^2+cx+d$ . If we multiply the two equation, $g(x)$ and $h(x)$ it will make an equation which is $f_n (x)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$ and also $a+c=0$ $b+d+ac=0$ $ad+bc=0$ $bd=n$ .

by $a+c=0$ we can conclude that $a=-c$ . By substituting $a=-c$ to $ad+bc=0$ we will have, $bc-dc=c(b-d)=0$ and by that we will have 2 possible outcomes $b-d=0$ and $c=0$ .

By the first possible outcome $b-d=0$ we can conclude that $b=d$ and $bd=b^2=n$ and $n<1000$ so $b<31$ . if we subtitute $b=d$ and $a=-c$ to $b+d+ac=0$ , we will have $2b-c^2=0$ , so $c=sqrt{2b}$ this means $2b$ is a perfect square to make $c$ an integer. The only values for $b$ is ${2,8,18}$ which give $n$ have the values ${4,64,324}$

By the second possible outcome $c=0$ , if substituted to $b+d+ac=0$ we will have $b+d=0$ so $b=-d$ but it cannot be because n must be positive and by $b=-d$ this makes $bd$ to be negative so we disregard $c=0$ as a possible out come.

So the only values for $n$ are ${4,64,324}$ , the sum of them is $4+64+324 = 392$ . The last three digits is $392$

According to the question, f(x)=g(x)*h(x) where g(x) and h(x) are non-constant polynomials.

It is obvious that the leading coefficients of g(x) and h(x) must be 1 or -1 each.

If for some n $x^4+n$ can be expressed as a product of non-constant polynomials g(x) and h(x) with leading coefficients -1 then $x^4+n$ can also be expressed as a product of 2 non-constant polynomials having leading coefficients 1(This can be achieved by simply multiplying g(x) and h(x) each by -1).

So, we are left with 2 cases.

$Case 1$

Let $x^4+n=(x+a)(x^3+bx^2+cx+n/a)$ where a,b,c are integers and $a|n$ .

Comparing coefficients on both sides we can see that this is not possible for +ve integer n.

$Case 2$

Let $x^4+n=(x^2+ax+b)(x^2+cx+n/b)$ where a,b,c are integers and b|n.

Comparing coefficients we get $b=\frac{a^2}{2}$ $c=-a$ $n=\frac{a^4}{4}$

This means n must be even.

Setting $\frac{a^4}{4}<1000$ we get a=2,4,6.

Therefore the required answer is $\boxed{392}$ .

We can express x^4+n as (x^4+2 sqrt(n) x^2+n)-2 sqrt(n) x^2=(x^2+sqrt(n))^2-(sqrt 4 x)^2=(x^2+sqrt 4 x+sqrt(n)(x^2-sqrt 4 *x+sqrt(n)). For the coefficients of these factors to be integers, we must have sqrt(n) be an integer, so n must be a perfect square. We let n=k^2. The fourth root of 4n=4k^2 is equal to sqrt(2k). Thus 2k must be a perfect square as well. We let k=2m^2 and have n=4m^4. When m=1, 2, 3, 4, we have n=4, 64, 324, 1024, respectively. Thus, the answer is 4+64+324=392.

For a number $\sqrt[4]{k}c, \; f_{kc^4}(x) = x^4 + kc^4$

$x^4 + kc^4 = (x^2+axc+\sqrt{k}c^2)(x^2-axc+\sqrt{k}c^2)$

From this we see that $a^2 = 2\sqrt{k}$

$kc^4 \in \mathbb{Z}^+ \Rightarrow \left \{k,c \right \} \subset \mathbb{Z}^+ \; so \; a \in \mathbb{Z}^+$

$kc^4 < 1000 \Rightarrow k<1000 \Rightarrow \sqrt{k} \leq 31$

$\therefore a^2 \leq 62 \; and \; \because 2\sqrt{k} = a^2, \; a$ is even.

*The square of any even number is even and the square of any odd number is odd, hence all even square numbers are squares of even numbers.

$a \in \left \{2,4,6 \right \} \Rightarrow \frac {a^2}{2} \in \left \{2,8,18 \right \}$

So $\sqrt{k} \in \left \{2,8,18 \right \}$ hence $k \in \left \{4,64,324 \right \}$

$c =1 \Rightarrow n=k,\; n \in \left \{4,64,324 \right \}$

$c = 2 \Rightarrow n = 16k, \; n \in \left \{ 64, 1024, 5184 \right \}$

$c = 3 \Rightarrow n =81k, \; n \in \left \{ 324,5184,26244 \right \}$

Obviously $1 \leq c \leq 3$ So the only possible values of $n$ are $4,64$ and $324$

$4+64+324 =392$

Note that if the polynomial $f$ will always be greater than 0 for $n > 0$ . Thus, the polynomial has no real roots, which means that the polynomial is the product of two degree 2 polynomials (product of degree 1 and degree 3 is eliminated).

Let the degree 2 polynomials be $x^2+ax+b$ and $x^2+cx+d$ . Note that these polynomials have leading coefficient 1, which is because the leading coefficient of $f$ is 1 and that the two polynomials must have integer coefficients. Multiplying, we obtain $x^4 + (a+c)x^3 + (b+ac+d)x^2 + (bc+ad)x + bd$ .

Then we obtain the equations: $\\ a+c=0 \\ b+ac+d = 0 \\ bc+ad = 0 \\ bd=n \\$ .

From the first and third equations, we can deduce that $a(b-d) = 0$ (multiplying the first equation by b and subtracting the third equation from this new one). If $a=0$ , then $a+c = 0 \implies c=0$ , and $b+ac+d = 0 \implies b=-d$ . However, this is impossible because $bd=-b^2=n$ , where $n > 0$ . Thus $b-d = 0 \implies b=d$ .

Now, consider the first and second original equations. We have $a+c=0 \implies a=-c$ , and that $b+ac+d = 0 \implies 2b-a^2 = 0 \implies 2b=a^2$ . Finding all such $b$ and summing up over the squares will obtain the answer, because all the equations will be met in this process. Luckily, this is not a terribly long casework, as $b=a^2/2$ , which means a is even because $b^2$ is an integer. Furthermore, $b^2<1000 \implies b<32$ , so $b=2^2/2, 4^2/2, \text{and } 6^2/2$ ( $8^2/2 =32$ ). We have $b=2, 8, 18$ . Then the answer is $2^2+8^2+18^2 = \boxed{392}$ .

we can write fn(x)=x^4+n as fn(x)=x^4+4y^4 or fn(x)=x^4+(y^4)/4 for integer n<1000

for n=4.y^4

y=1 => n=4.1^4=4

y=2 => n=4.2^4=64

y=3 => n=4.3^4=324

y=4 => n=4.4^4=1024 outer limit

and

for n=(1/4)y^4

y=1 => n=(1/4)1^4=1/4 outer limit

y=2 => n=(1/4).2^4=4

y=4 => n=(1/4).4^4=64

y=4 => n=(1/4).6^4=324

y=8 => n=(1/4).8^4=1024 outer limit

so sum of all n= 4+64+324=392

I know this is cheating but the cheating might be justified. We know that the expression (a^4 + 4 b^4) can be factorized as [(a + b)^2 + b^2] * [(a - b)^2 + b^2]. Considering this as the only factorization possible for positive n, we get n=4 b^4 for "b" belonging to positive integers. now n<1000. therefore maximum value of b is 3. so we get n = 4, 64 and 324, the sum of which is 392.

We know that polynomials of the form $x^{4} + 4y^{4}$ can be factored into two polynomials with integer coefficients, namely, $(x^{2} + 2xy + 2y^{2}) (x^{2} - 2xy + 2y^{2} )$ .

Thus n can be $4 \times 1^{4}$ , $4 \times 2^{4}$ and $4 \times 3^{4}$ as $4 \times 4^{4}$ exceeds 1000.

Hence the required answer is $4 + 64 + 324$ = $\boxed{392}$