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If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$

I have tried putting

\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}

Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}

Somehow it does not match the statement given.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Crocogator
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4 Answers4

7

You correctly computed $$ (\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2 $$ only that this is not what was asked for.

But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore $$ (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i) $$ which expands to $$ (1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, . $$

Martin R
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3

Denoting $\alpha=x_1$, $\beta=x_2$, $\gamma=x_3$ and $\delta=x_4$, we have $$\small\prod_{i=1}^4(1+x_i^2)-\prod_{i=1}^4(1-x_i^2)=\small2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)+2(x_1+x_2+x_3+x_4)$$ so taking out a factor of $x_1x_2x_3x_4$ from the first term in brackets, $$\prod_{i=1}^4(1+x_i^2)=2\left(\prod_{i=1}^4x_i\right)\left(\sum_{i=1}^4\frac1{x_i}\right)+2\sum_{i=1}^4x_i+\prod_{i=1}^4(1-x_i^2)$$ which is evaluable by Vieta's formulas. For the sum of reciprocals, see this post.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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    Sorry but I can't make heads or tails of it. Looking at the post it seems like differentiation is involved somewhere. And can you please tell me where to start to understand this "Vieta's formulas" coz I'm a newbie to Advance math like these. – Crocogator Oct 12 '19 at 14:03
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    Added link to the formulas. A lot of sums and products, but no calculus involved, as per the tags of your post. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Oct 12 '19 at 14:06
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    Oh Thanks ^_^ I'll go through the links. I just realized my book also has it. (That's where I found this question). This formula is given in the later articles on the book. – Crocogator Oct 12 '19 at 14:07
2

Another way using Transformation of equation:

We have $$x^4+qx^2+s=-x(px^2+r)$$

Squaring both sides

$$(x^2)^4+q^2(x^2)+s^2+2q(x^2)^3+2qs(x^2)+2s(x^2)^2=x^2(p^2(x^2)^2+r^2+2pr(x^2))$$

Replacing $x^2+1=y$

$$(y-1)^4+q^2(y-1)^2+s^2+2q(y-1)^3+2qs(y-1)+2s(y-1)^2=(y-1)(p^2(y-1)^2+r^2+2pr(y-1))$$

$$\iff y^4+\cdots+1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr=0$$

Using Vieta's formula $$\prod_{j=1}^4y_j=\dfrac{1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr}1=(1-q+s)^2+(p-r)^2$$

lab bhattacharjee
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-2

Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.

Alan Muniz
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Chirag Singla
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