a problem on complex numbers












2














Let $wneq 1$ and $w^{13} = 1$.



If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?



I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?






complex-numbers







share|cite|improve this question
























  • Welcome to the site. Please typeset your equations with Mathjax for better presentation.
    – Shubham Johri
    2 days ago






  • 1




    These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
    – Lord Shark the Unknown
    2 days ago








  • 1




    Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
    – K Split X
    2 days ago








  • 2




    "And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
    – fleablood
    2 days ago










  • Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
    – rschwieb
    2 days ago


















2














Let $wneq 1$ and $w^{13} = 1$.



If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?



I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?






complex-numbers







share|cite|improve this question
























  • Welcome to the site. Please typeset your equations with Mathjax for better presentation.
    – Shubham Johri
    2 days ago






  • 1




    These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
    – Lord Shark the Unknown
    2 days ago








  • 1




    Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
    – K Split X
    2 days ago








  • 2




    "And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
    – fleablood
    2 days ago










  • Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
    – rschwieb
    2 days ago
















2












2








2







Let $wneq 1$ and $w^{13} = 1$.



If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?



I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?






complex-numbers







share|cite|improve this question















Let $wneq 1$ and $w^{13} = 1$.



If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?



I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Andrei

11.3k21026




11.3k21026










asked 2 days ago









Rituraj Tripathy

112




112












  • Welcome to the site. Please typeset your equations with Mathjax for better presentation.
    – Shubham Johri
    2 days ago






  • 1




    These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
    – Lord Shark the Unknown
    2 days ago








  • 1




    Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
    – K Split X
    2 days ago








  • 2




    "And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
    – fleablood
    2 days ago










  • Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
    – rschwieb
    2 days ago




















  • Welcome to the site. Please typeset your equations with Mathjax for better presentation.
    – Shubham Johri
    2 days ago






  • 1




    These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
    – Lord Shark the Unknown
    2 days ago








  • 1




    Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
    – K Split X
    2 days ago








  • 2




    "And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
    – fleablood
    2 days ago










  • Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
    – rschwieb
    2 days ago


















Welcome to the site. Please typeset your equations with Mathjax for better presentation.
– Shubham Johri
2 days ago




Welcome to the site. Please typeset your equations with Mathjax for better presentation.
– Shubham Johri
2 days ago




1




1




These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
– Lord Shark the Unknown
2 days ago






These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
– Lord Shark the Unknown
2 days ago






1




1




Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
– K Split X
2 days ago






Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
– K Split X
2 days ago






2




2




"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
– fleablood
2 days ago




"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
– fleablood
2 days ago












Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
– rschwieb
2 days ago






Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
– rschwieb
2 days ago












2 Answers
2






active

oldest

votes


















2














Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$



Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$






share|cite|improve this answer























  • I got to this point. I can not solve the trigonometric part.
    – Rituraj Tripathy
    2 days ago










  • I went back to $w$ and wrote explicitly the product
    – Andrei
    2 days ago










  • I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
    – gandalf61
    2 days ago












  • Fixed that. I forgot to add and subtract one. I've just added.
    – Andrei
    2 days ago










  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
    – Will Jagy
    yesterday



















2














$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$



This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore



$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$



and
$$ a^2 + a - 3 = 0 $$






share|cite|improve this answer





















  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
    – Andrei
    yesterday











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$



Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$






share|cite|improve this answer























  • I got to this point. I can not solve the trigonometric part.
    – Rituraj Tripathy
    2 days ago










  • I went back to $w$ and wrote explicitly the product
    – Andrei
    2 days ago










  • I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
    – gandalf61
    2 days ago












  • Fixed that. I forgot to add and subtract one. I've just added.
    – Andrei
    2 days ago










  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
    – Will Jagy
    yesterday
















2














Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$



Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$






share|cite|improve this answer























  • I got to this point. I can not solve the trigonometric part.
    – Rituraj Tripathy
    2 days ago










  • I went back to $w$ and wrote explicitly the product
    – Andrei
    2 days ago










  • I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
    – gandalf61
    2 days ago












  • Fixed that. I forgot to add and subtract one. I've just added.
    – Andrei
    2 days ago










  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
    – Will Jagy
    yesterday














2












2








2






Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$



Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$






share|cite|improve this answer














Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$



Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Andrei

11.3k21026




11.3k21026












  • I got to this point. I can not solve the trigonometric part.
    – Rituraj Tripathy
    2 days ago










  • I went back to $w$ and wrote explicitly the product
    – Andrei
    2 days ago










  • I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
    – gandalf61
    2 days ago












  • Fixed that. I forgot to add and subtract one. I've just added.
    – Andrei
    2 days ago










  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
    – Will Jagy
    yesterday


















  • I got to this point. I can not solve the trigonometric part.
    – Rituraj Tripathy
    2 days ago










  • I went back to $w$ and wrote explicitly the product
    – Andrei
    2 days ago










  • I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
    – gandalf61
    2 days ago












  • Fixed that. I forgot to add and subtract one. I've just added.
    – Andrei
    2 days ago










  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
    – Will Jagy
    yesterday
















I got to this point. I can not solve the trigonometric part.
– Rituraj Tripathy
2 days ago




I got to this point. I can not solve the trigonometric part.
– Rituraj Tripathy
2 days ago












I went back to $w$ and wrote explicitly the product
– Andrei
2 days ago




I went back to $w$ and wrote explicitly the product
– Andrei
2 days ago












I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
– gandalf61
2 days ago






I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
– gandalf61
2 days ago














Fixed that. I forgot to add and subtract one. I've just added.
– Andrei
2 days ago




Fixed that. I forgot to add and subtract one. I've just added.
– Andrei
2 days ago












Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
– Will Jagy
yesterday




Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
– Will Jagy
yesterday











2














$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$



This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore



$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$



and
$$ a^2 + a - 3 = 0 $$






share|cite|improve this answer





















  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
    – Andrei
    yesterday
















2














$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$



This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore



$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$



and
$$ a^2 + a - 3 = 0 $$






share|cite|improve this answer





















  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
    – Andrei
    yesterday














2












2








2






$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$



This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore



$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$



and
$$ a^2 + a - 3 = 0 $$






share|cite|improve this answer












$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$



This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore



$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$



$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$



and
$$ a^2 + a - 3 = 0 $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Will Jagy

101k599199




101k599199












  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
    – Andrei
    yesterday


















  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
    – Andrei
    yesterday
















Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
– Andrei
yesterday




Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
– Andrei
yesterday


















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