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let sequence $\{b_{n}\}$ such $$b_{n}=-b_{n-1}-2b_{n-2}(n\ge 3),b_{1}=1,b_{2}=-1$$ show that $$ 2b_k^4+8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4=0$$

My try:since $$b_{k}=-b_{k-1}-2b_{k-2}$$ so $$2(-b_{k-1}-2b_{k-2})^4+24(-b_{k-1}-2b_{k-2})^2b^2_{k-1}+32(-b_{k-1}-2b_{k-2})b^3_{k-1}+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4\neq 0$$

see http://www.wolframalpha.com/input/?i=2%28x%2B2y%29%5E4%2B24%28x%2B2y%29%5E2*x%5E2%2B32%28-x-2y%29*x%5E3%2B14x%5E4-48x%5E2*y%5E2-32y%5E3

My methods is wrong? so How can prove it? Thank you

and By the way this problem is from this:Does this sequence always give a square number?

But I find this red part maybe is not equals zero,(so this proof is not complete)

Community
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math110
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2 Answers2

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$2b_k^4+8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4$

$ = (2b_k^4+8b_k^3b_{k-1}+12b_k^2b_{k-1}^2+8b_kb_{k-1}^3+2b_{k-1}^4)+12b_k^2b_{k-1}^2+24b_kb_{k-1}^3+12b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4$

$= 2(b_k+b_{k-1})^4+12(b_k^2b_{k-1}^2+2b_kb_{k-1}^3+b_{k-1}^4)-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4$

$= 32b_{k-2}^4+12b_{k-1}^2(-2b_{k-2})^2-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4$

$= 0$

r9m
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Given $$ x=b_k, \; \; y = b_{k-1}, \; \; z = b_{k-2}, $$ your polynomial is double $$ \left( x^2 + 2 x y + 4 y^2 \right)^2 - \left( 3 y^2 + 4 z^2 \right)^2 $$ However, $$ x = -y - 2 z. $$ So $$ x^2 + 2 x y + 4 y^2 = (x+y) + 3 y^2 = (-2z)^2 + 3 y^2 = 4 z^2 + 3 y^2, $$ and this equals $$ 3y^2 + 4 z^2. $$ The squaring does not change the equality.

Will Jagy
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