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Determine all positive integers \(k\le 2000\)  for which \(x^4 + k\)  can be factored into two distinct trinomial factors with integer coefficients.

Guest Jul 25, 2018
 #1
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This is the updated question of what I made before

Guest Jul 25, 2018
 #2
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Personal opinion: Of the people who have looked at your problem, nobody seems to be able to solve it so far. It is not because nobody wants to help you. The problem seems genuinely to be a difficult one.

Guest Jul 25, 2018
 #3
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Suppose x4+k=(a1*x2+b1*x+c1)*(a2*x2+b2*x+c2). this means that a1*a2*x4=x4, therefore a1*a2=1. to make things simpler, we can multiply the first trinomial by a2 and the second trinomial by a1 (this doesn't change the result, because a1*a2=1) and get:

 

x4+k=(a2*a1*x2+a2*b1*x+a2*c1)*(a1*a2*x2+a1*b2*x+a1*c2)=(1*x2+a2*b1*x+a2*c1)*(1*x2+a1*b2*x+a1*c2​)=

 

(x2+a2*b1*x+a2*c1)*(x2+a1*b2*x+a1*c2​) 

 

(I did this because i wanted to get rid of the coefficients of x2 in the trinomials. we can now conclude that x4+k is a product of two trinomials with integer coefficients if and only if  it is a product of two trinomials with integer coefficients where the coefficient of x2 in each of the trinomials is 1. Now we can assume that the coefficients of x2 in each of the trinomials)

 

suppose x4+k=(x2+a*x+b)*(x2+c*x+d)=x4+a*x3+c*x3+b*x2+d*x2+(a*c)*x2+(a*d)*x+(c*b)*x+b*d=

 

x4+x3*(a+c)+x2*(b+d+a*c)+x*(a*d+c*b)+b*d. From this we can conclude that a+c=0, meaning that c=-a.

 

Let's substitute c with -a:

 

x4+k=x4+x3*(a+(-a))+x2*(b+d+a*(-a))+x*(a*d+(-a)*b)+b*d=x4+x2*(b+d-a2)+x*(a*(d-b))+b*d

 

If a=0:

 

x4+k=x4+x2*(b+d-a2)+x*(a*(d-b))+b*d=x4+x2*(b+d)+b*d. the coefficient of x2 has to be 0, therefore d=-b. Let's substitute d with -b:

 

x4+k=x4+x2*(b+d)+b*d=x4+x2*0+b*(-b)=x4-b2. this means that k=-b2. but -b2 is a nonpositive number (because b2 is nonnegative) meaning that k has to be nonpositive. But k has to be positive, so that is a contradiction.

 

Therefore 'a' CANNOT be 0.

 

Now that we know we can divide by 'a':

 

x4+k=x4+x2*(b+d-a2)+x*(a*(d-b))+b*d. The coefficient of x is supposed to be 0, therefore a*(d-b)=0, meaning that d-b=0, and that d=b. Let's substitute d with b:

 

x4+k=x4+x2*(b+b-a2)+x*(a*(b-b))+b*b=x4+x2*(2*b-a2)+b2. the coefficient of x2 is supposed to be 0, meaning that 2*b-a2=0, and that 2*b=a2. this means that b=a2/2. We also know that k=b2, meaning that k=a4/4. This means that if an integer k satisifies the condition that is specified in the question, there exist a positive integer a such that a4/4=k. b=a2/2, meaning that 'a' has to be an even number. a=2*n, meaning that n4*4=k. so k satistifes the conditions if there exist a positive integer n such that n4*4=k. 14*4=4, 24*4=64, 34*4=324. We need to find the number of nonnegative integers n such that n4*4<=2000.

 

n4*4<=2000 if and only if n4<=500 if and only if n<=5001/4≈4.72870804502. Therefore, the largest positive integer n that satisfies n4*4<=2000 is 4, and there are exactly 4 positive integers that satisfy the equation- 1, 2, 3 and 4. This means that there are 4 values of k that are positive and <=2000 that have a positive integer n that satisfies n4*4<=2000-

 

k=14*4=4, k=24*4=64, k=34*4=324, and k=44*4=45=1024. So, There are exactly 4 values of k that satisfy the following conditions:

 

1. k is a positive integer between 1 and 2000

2. x4+k can be factored into 2 distinct trinomials, with integer coefficients

 

x4+4=(x2+2*x+2)*(x2-2*x+2)

 

x4+64=(x2+4*x+8)*(x2-4*x+8)

 

x4+324=(x2+6*x+18)*(x2-6*x+18)

 

x4+1024=(x2+8*x+32)*(x2-8*x+32)

Guest Jul 25, 2018
edited by Guest  Jul 25, 2018
edited by Guest  Jul 25, 2018
 #4
avatar+89953 
+2

Let us assume that  we can factor  x^4 + k  in the following manner :

 

(x^2  + ax  + b) (x^2 - ax  + b)

Expanding this, we have 

 

x^4 + ax^3 + bx^2

      -  ax^3 -a^2x^2 - abx

                  +bx^2   + abx   + b^2

___________________________

x^4      +   (2b - a^2)x^2         + b^2

 

Since the  x^2 term will "disappear", this implies that  b^2  = k

Then  b can possibly range from 1 - 44   since 1^2  = 1  and 44^2  = 1936.....note that 45^2  = 2025 which is too large

 

And since the  x^2 term disappears, it must be that  2b - a^2  =  0

Add a^2 to both sides  and we have that   2b  = a^2

Taking the square root of this we have that

√[2b]  = a

Since a is an integer, then  √[2b]  must also  be an integer

And the values that make √[2b]  an integer is when  b  = 2    b  = 8    b = 18   and b  = 32

 

So  (a, b)   =  ( 2,2)  (4, 8)   ( 6, 18)    and  ( 8, 32)

 

So........the following  product trinomials are possible

 

( x^2 + 2x + 2) ( x^2 - 2x + 2)  =  x^2 + 4

(x^2 + 4x + 8) (x^2 - 4x + 8) = x^2  + 64

(x^2 + 6x + 18) (x^2 - 6x + 18)   x^2 + 324

(x^2 + 8x  + 32) (x^2 - 8x + 32)  = x^2 + 1024

 

 

cool cool cool

CPhill  Jul 26, 2018
 #5
avatar+20025 
+1

Determine all positive integers
\(k \le 2000 \)  
for which
\(x^4+k\)
can be factored into two distinct trinomial factors with integer coefficients.

 

\(\text{All four roots of $x^4+k$ are complex numbers, because $k$ is a positive integer} \\ \text{The complex numbers are conjugation in pairs :} \)

 

\(\begin{array}{|rcll|} \hline x^4+k &=& (x-x_1)(x-x_2)(x-x_3)(x-x_4) \quad | \quad x_1,x_2,x_3,x_4 \in C \\ && x_1 = a+bi \\ && x_2 = a-bi \\ && x_3 = u+vi \\ && x_4 = u-vi \\ x^4+k &=& \Big(x-(a+bi)\Big) \Big(x-(a-bi)\Big) \Big(x-(u+vi)\Big) \Big(x-(u-vi)\Big) \\ &=& \Big((x-a)-bi\Big) \Big((x-a)+bi)\Big) \Big((x-u)-vi)\Big) \Big((x-u)+vi)\Big) \\ &=& \Big((x-a)^2-b^2i^2\Big) \Big((x-u)^2-v^2i^2)\Big) \quad | \quad i^2=-1\\ &=& \Big((x-a)^2+b^2\Big) \Big((x-u)^2+v^2)\Big) \\ &=& \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \quad | \quad \text{trinomial factors}\\ &=& x^4-2ux^3+(u^2+v^2)x^2-2ax^3+4aux^2-2ax(u^2+v^2) \\ && +(a^2+b^2)x^2-2u(a^2+b^2)x+(a^2+b^2)(u^2+v^2) \\ &=& x^4 -x^3\underbrace{(2u+2a)}_{=0} +x^2\underbrace{(u^2+v^2+a^2+b^2+4au)}_{=0} \\ && -x\underbrace{\Big(2a(u^2+v^2)+2u(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(u^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 1. & 2u+2a &=& 0 \quad & | \quad : 2 \\ & u+a &=& 0 \\ & \mathbf{u} &\mathbf{=}& \mathbf{-a} \quad & | \quad \text{or} \quad \mathbf{u^2=a^2} \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(a^2+v^2+a^2+b^2-4a^2)}_{=0} \\ && -x\underbrace{\Big(2a(a^2+v^2)-2a(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(v^2+b^2-2a^2)}_{=0} -x\underbrace{(2av^2-2ab^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 2. & 2av^2-2ab^2 &=& 0 \quad & | \quad : 2a \\ & v^2-b^2 &=& 0 \\ & \mathbf{v^2 } &\mathbf{=}& \mathbf{b^2 } \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(b^2+b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(2b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 3. & 2b^2-2a^2 &=& 0 \quad & | \quad : 2 \\ & b^2-a^2 &=& 0 \\ & \mathbf{b^2 } &\mathbf{=}& \mathbf{a^2 } \\ \hline \end{array}\\ &=& x^4 +\underbrace{(a^2+a^2)(a^2+a^2)}_{=k} \\ &=& x^4 +\underbrace{(2a^2)(2a^2)}_{=k} \\ &=& x^4 +\underbrace{4a^4}_{=k} \\ \hline && \huge{k=4a^4} \\ \hline \end{array}\)

 

\(\text{trinomial factors:}\)

\(\begin{array}{|rcll|} \hline && \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \\ && \boxed{ b^2 = a^2, \quad u^2 = a^2, \quad v^2=b^2=a^2, \quad u = -a } \\ &=& ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline \end{array} \)

 

Solutions:

\(\begin{array}{|r|r|c|c|} \hline a \in N &k= 4a^4 & k \le 2000 & ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline 1 & 4 & \checkmark & ( x^2-2x+ 2)(x^2+2x + 2) \\ 2 & 64 & \checkmark & ( x^2-4x+ 8)(x^2+4x + 8) \\ 3 & 324 & \checkmark & ( x^2-6x+ 18)(x^2+6x + 18) \\ 4 & 1024 & \checkmark & ( x^2-8x+ 32)(x^2+8x + 32) \\ 5 & 2500 & \gt 2000 \text{ no solution } \\ \ldots & \ldots & \gt 2000 \text{ no solution } \\ \hline \end{array}\)

 

All positive integers k  are 1,2,3,4

 

laugh

heureka  Jul 27, 2018
edited by heureka  Jul 27, 2018

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