Algebraic Techniques 02 - Factorization

• ✔ Sophie-Germain’s Identity
• ✔ Useful Identities to Remember
• Factorize $\, x^4 + x^2 + 1$.
• Factorize $\, x^5 + x + 1$.
• Given the following equations for all real numbers $\, a$, $b$, $c$, and $d$, find $\, d + a + 2da$.
• For all real numbers $\, a$, $b$, $c$, and $d$ such that $\, a^2 + b^2 + c^2 + d^2 \leq 1$, find the maximum of the following expression.
• For real numbers $\, a$, $b$, and $c$, prove the following equation.
• Reference

✔ Sophie-Germain’s Identity

$$\begin{aligned} x^4 + 4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2) \end{aligned}$$

✔ Useful Identities to Remember

$$\begin{aligned} a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \end{aligned}$$ $$\begin{aligned} a^2 + b^2 + c^2 - ab - bc - ca = \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2} \end{aligned}$$ $$\begin{aligned} a^2 + b^2 + c^2 + ab + bc + ca = \frac{(a + b)^2 + (b + c)^2 + (c + a)^2}{2} \end{aligned}$$

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Factorize $\, x^4 + x^2 + 1$.

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[Solution] Adding and subtracting $x^2$, $$\begin{aligned} x^4 + x^2 + 1 = x^4 + 2x^2 + 1 - x^2 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1) \end{aligned}$$

Another solution is to use the technique in the previous post $$\begin{aligned} x^4 + x^2 + 1 &= (x^2 + a)^2 - (bx + c)^2 \\ &= x^4 + (2a - b^2)x^2 - 2bcx + a^2 - c^2 \end{aligned}$$

Comparing coefficients, $2a - b^2 = 1$, $-2bc = 0$, and $a^2 - c^2 = 1$. So $b = 0$ or $c = 0$. If $b = 0$, it makes $a = 1/2$ and there is a contradiction getting $c^2 + 3/4 = 0$. So $c = 0$ and $a^2 = 1$. Finally, $a = 1$ and $b^2 = 1$ since another contradiction comes up when $a = -1$, which allows the same result. $$\begin{aligned} x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1) \end{aligned}$$

Factorize $\, x^5 + x + 1$.

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[Solution] Adding and subtracting $x^2$, $$\begin{aligned} x^5 + x + 1 = x^5 - x^2 + x^2 + x + 1 = x^2 (x^3 - 1) + x^2 + x + 1 \end{aligned}$$

Considering that $x^3 - 1 = (x - 1)(x^2 + x + 1)$, $$\begin{aligned} x^2 (x^3 - 1) + x^2 + x + 1 = (x^2 + x + 1)(x^2 (x - 1) + 1) = (x^2 + x + 1)(x^3 - x^2 + 1) \end{aligned}$$

Given the following equations for all real numbers $\, a$, $b$, $c$, and $d$, find $\, d + a + 2da$.

$$\begin{aligned} a + b + 2ab = 3, \quad b + c + 2bc = 4, \quad c + d + 2cd = -5 \end{aligned}$$

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[Solution] It is important to observe the patterns of these equations. Using the fact $$\begin{aligned} 1 + 2(x + y + 2xy) = 1 + 2x + 2y + 4xy = (1 + 2x)(1 + 2y) \end{aligned}$$

these equations can be factorized as $$\begin{aligned} (1 + 2a)(1 + 2b) = 7, \quad (1 + 2b)(1 + 2c) = 9, \quad (1 + 2c)(1 + 2d) = -9 \end{aligned}$$

Let $x = 2a + 1$, $y = 2b + 1$, $z = 2c + 1$, and $t = 2d + 1$, then $xy = 7$, $yz = 9$, and $zt = -9$. So, $y = 7/x$, $z = 9/y = 9x/7$, and $(9x/7)t = -9$. It implies that $xt = -7$. $$\begin{aligned} d + a + 2da = \frac{xt - 1}{2} = -4 \end{aligned}$$

For all real numbers $\, a$, $b$, $c$, and $d$ such that $\, a^2 + b^2 + c^2 + d^2 \leq 1$, find the maximum of the following expression.

$$\begin{aligned} (a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \end{aligned}$$

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[Solution] The main idea that is hard to come up with is adding $(a - b)^4$ to $(a + b)^4$. $$\begin{aligned} \color{crimson}{(a + b)^4 + (a - b)^4} &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 + a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 \\ &= \color{crimson}{2(a^4 + b^4 + 6a^2b^2)} \end{aligned}$$

Using this fact, $$\begin{aligned} &(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \\ + \, &(a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 \\ = \, &6(a^4 + b^4 + c^4 + d^4) + 12(a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2 + a^2c^2 + b^2d^2) \end{aligned}$$

Considering $(x + y + z + t)^2 = x^2 + y^2 + z^2 + t^2 + 2(xy + yz + zt + tx + xz + yt)$, the right side of the above equation is $6(a^2 + b^2 + c^2 + d^2)^2$. By the constraint of the problem, $6(a^2 + b^2 + c^2 + d^2)^2 \leq 6$. Therefore, $$\begin{aligned} (a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \leq 6 \end{aligned}$$

The equal sign is valid when $a = b = c = d = \pm 1 / 2$.

For real numbers $\, a$, $b$, and $c$, prove the following equation.

$$\begin{aligned} (a - b)^5 + (b - c)^5 + (c - a)^5 = 5(a - b)(b - c)(c - a)(a^2 + b^2 + c^2 - ab - bc - ca) \end{aligned}$$

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[Solution] Let $x = a - b$, $y = b - c$, and $z = c - a$, then $x + y + z = 0$ and $$\begin{aligned} a^2 + b^2 + c^2 - ab - bc - ca = \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2} = \frac{x^2 + y^2 + z^2}{2} \end{aligned}$$

So, it is sufficient to prove that $$\begin{aligned} x^5 + y^5 + z^5 = \frac{5}{2} xyz (x^2 + y^2 + z^2) \end{aligned}$$

By $z = -x - y$, $$\begin{aligned} x^5 + y^5 + z^5 &= x^5 + y^5 - (x + y)^5 \\ &= (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4 - (x + y)^4) \\ &= (x + y)(-x^3y + x^2y^2 - xy^3 - 4x^3y - 6x^2y^2 - 4xy^3) \\ &= -5(x + y)(x^3y + xy^3 + x^2y^2) \\ &= 5zxy(x^2 + y^2 + xy) \end{aligned}$$

Since $$\begin{aligned} \frac{5}{2} xyz (x^2 + y^2 + z^2) = \frac{5}{2} xyz (x^2 + y^2 + (x + y)^2) = 5xyz(x^2 + xy + y^2) \end{aligned}$$

comparing two equations completes the proof. Another way to prove this is by using relationship between roots and coefficients of cubic equation. Let $S = xy + yz + zx$ and $P = xyz$, then for any real number $t$, $$\begin{aligned} (t - x)(t - y)(t - z) = t^3 + St - P \end{aligned}$$

Since $x$, $y$, and $z$ are the solutions of $t^3 + St - P = 0$, so does $t^5 + St^3 - Pt^2 = 0$. Adding two equations, $x$, $y$, and $z$ are also the solutions of $t^5 + S(P - St) - Pt^2 = 0$. It shows that $$\begin{aligned} x^5 + y^5 + z^5 + S(3P - S(x + y + z)) - P(x^2 + y^2 + z^2) = 0 \end{aligned}$$

Using the fact that $x + y + z = 0$, $$\begin{aligned} & x^5 + y^5 + z^5 + 3SP - P(x^2 + y^2 + z^2) = 0 \\ \Longleftrightarrow \, & x^5 + y^5 + z^5 = P(x^2 + y^2 + z^2 - 3S) \end{aligned}$$

So, it is sufficient to prove that $$\begin{aligned} x^2 + y^2 + z^2 - 3(xy + yz + zx) = \frac{5}{2} (x^2 + y^2 + z^2) \end{aligned}$$

It implies that $x^2 + y^2 + z^2 + 2(xy + yz + zx) = (x + y + z)^2 = 0$, which is obvious because $x + y + z = 0$.

Reference

[1] Titu Andreescu, 105 Algebra Problems.