Problems

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Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.

1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.

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Solution:

1. \begin{align*} && 0 &= \alpha^2 + a \alpha + b \tag{1} \\ && 0 &= \alpha^2 + c \alpha + d \tag{2} \\ \\ (1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\ \Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)} \end{align*} (\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have: \begin{align*} && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \end{align*} (\(\Leftarrow\)) Suppose the equation holds, then \begin{align*} && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\ \Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \end{align*} So \(\alpha\) is a root of the first equation. Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations. If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
2. \begin{align*} && 0 &= x^2+ax+b \tag{1} \\ && 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\ \\ (2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3} \end{align*} Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition. \(a = \frac52, q = \frac52, r = \frac12\) \begin{align*} && 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\ &&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\ &&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\ &&&= (b-1)(4b^2-2b-1) \\ \Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}

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Solution: \(p(x) = q(x)(x-1)^5 + 1\) where \(q(x)\) has degree \(4\).

1. \(p(1) = q(1)(1-1)^5 + 1 = 1\).
2. \(p'(x) = q'(x)(x-1)^5 + 5(x-1)^4q(x) + 0 = (x-1)^4((x-1)q'(x) + 5q(x))\) so \(p'(x)\) is divisible by \((x-1)^4\)
3. \(p(x)+1\) divisible by \((x+1)^5\) means that \(p(-1) = -1\) and \(p'(x)\) is divisible by \((x+1)^4\). Since \(p'(x)\) is degree \(8\) it must be \(c(x+1)^4(x-1)^4 = c(x^2 - 1)^4\). Expanding and integrating, we get \(p(x) = c(\frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + d\). When \(x = 1\) we get \(c \frac{128}{315} + d = 1\) and when \(x = -1\) we get \(-c \frac{128}{315} + d = -1\) so \(2d = 0 \Rightarrow d = 0, c = \frac{315}{128}\) and \[ p(x) =\frac{315}{128} \l \frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x\r \]

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Solution:

1. If \(n = 1\) the theorem is \(\max_{x \in [-1,1]} \left ( |x + a_0 |\right) \geq 1\), but clearly \(\max(1+a_0, |a_0 - 1|) \geq 1\) (taking according to the sign of \(a_0\))
   1. \( \p(x) = x^2 - \frac12\,\) - take \(x = 0\) then \(|p(0)| = \frac12 \geq 2^{1-2} = \frac12\)
   2. \(\p(x) = x^3 - x \,\). take \(x = \frac1{\sqrt{2}}\), then \(|p\left ( \frac1{\sqrt{2}}\right)| = |\frac12 \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}| = \frac{1}{2\sqrt{2}} > \frac14 = 2^{1-3} \)
2. Consider \(p(x) = \frac{1}{64} \left ( 64x^5+25x^4-66x^3-24x^2+3x+1\right)\), then \(p\) satisfies the conditions of the theorem, therefore \(\max |p(x)| \geq 2^{1-5} = \frac1{16} = \frac{4}{64}\). However, \(p(-1) = \frac{1}{64}\) and \(p(1) = \frac{3}{64}\), so it cannot be strictly increasing or decreasing and there must be at turning point to achieve \(\frac{4}{64}\)

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Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.

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Solution:

1. \(\,\) \begin{align*} && x^2-y^2 &=(x-y)^3 \\ \Rightarrow && x+y &=d^2 \\ && x-y &= d \\ \Rightarrow && x &= \tfrac12(d^2+d) \\ && y &= \tfrac12(d^2-d) \end{align*} Therefore consider \(x^2 = m, y^2 = n\), so \(m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2\) so we want \(d^2-d > 20\), so \(d = 6, n = 225, m = 441\).
2. \(\,\) \begin{align*} && x^3-y^3 &= (x-y)^4 \\ \Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\ && d^3 &= (x-y)^2+3xy \\ && d^3 &= d^2 + 3xy \\ \Rightarrow && 3xy &= d^3 - d^2 \\ \Rightarrow && 3x(x-d) &= d^3-d^2 \\ \Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\ \Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\ &&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\ &&&= d \pm d \sqrt{\frac{4d-1}{3}} \end{align*} Therefore we need \(\frac{4d-1}{3}\) to be an odd square. \(y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}\). Since we want positive values, we should take the positive square roots. \(d = \frac{3 \cdot 3^2 + 1}{4} = 7\) we have \(2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7\)

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Solution: \begin{align*} && y &= ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) \\ && y(2) &= 8a-24a+24a+24-8a-16 \\ &&&= 8 \\ && y'(x) &= 3ax^2-12ax+(12a+12) \\ && y'(0) &= 12a-24a+12a+12 \\ &&&= 12 \end{align*} Therefore since our curve has the same value and gradient at \((2,8)\) as \(y = x^3\) they must touch at this point. Therefore \begin{align*} && ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) - x^3 &= (x-2)^2((a-1)x-(2a+4)) \end{align*} Therefore if \(a \neq 1\), they touch again when \(x = \frac{2a+4}{a-1}\).

1. 

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   + - ↺

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Solution: Claim: If \(f\) is a constant, then \(\triangle f = 0\) Proof: First consider \(f(x) = 1, g(x) = x\) then we must have: \begin{align*} && \triangle (1x) &= 1 \triangle x + x \triangle 1 \tag{iv} \\ &&&= 1 \cdot 1 + x \triangle 1 \tag{i} \\ \Rightarrow && 1 &= 1 + x \triangle 1 \tag{i} \\ \Rightarrow && \triangle 1 &= 0 \\ \Rightarrow && \triangle c &= 0 \tag{iii} \end{align*} \begin{align*} && \triangle (x^2) &= x \triangle x + x \triangle x \tag{iv} \\ &&&= x \cdot 1 + x \cdot 1 \tag{i} \\ &&&= 2x \\ \\ && \triangle (x^3) &= x^2 \triangle x + x \triangle (x^2) \tag{iv} \\ &&&= x^2 \cdot 1 + x \cdot 2x \tag{\(\triangle x^2 = 2x\)}\\ &&&= 3x^2 \end{align*} Claim: \(\triangle h(x) = \frac{\d h(x)}{\d x}\) for any polynomial \(h\) Proof: Since both \(\triangle\) and \(\frac{\d}{\d x}\) are linear (properties \((ii)\) and \((iii)\)) it suffices to prove that: \(\triangle x^n = nx^{n-1}\). For this we proceed by induction. Base cases (we've proved up to \(n = 3\) so we're good). Suppose it's true for some \(n\), then consider \(n + 1\), \begin{align*} && \triangle (x^{n+1}) &= x \triangle (x^n) + x^n \triangle x \tag{iv} \\ &&&= x \cdot n x^{n-1} + x^n \triangle x \tag{Ind. hyp.} \\ &&&= nx^n + x^n \tag{i} \\ &&&= (n+1)x^{n} \end{align*} Therefore it's true for for \(n+1\). Therefore by induction it's true for all \(n\).

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Solution:

1. Let \(f(x,y) = x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y\), then \begin{align*} f(3,y) &= 27 - 5 \cdot 9 +18y + 3y^2-24y-3y^2+18 + 6y \\ &= 0 \end{align*}, therefore \(x-3\) is a factor of \(f(x,y)\). \begin{align*} f(x,y) &= x^3-5x^2+6x+y(2x^2-8x+6) + y^2(x-3) \\ &= (x-3)(x^2-2x)+y(x-3)(2x-2)+y^2(x-3) \\ &= (x-3)(x^2-2x+2y(x-1)+y^2) \\ &= (x-3)(x+y)(x+y-2) \end{align*}
2. Let \(g(x,y) = 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\), notice that \(g(x,-2) = 0\), so \(y+2\) is a factor, \begin{align*} g(x,y) &= 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10 \\ &= x^2(2+y) + x(12-4y-5y^2) + 6y^3-y^2-21y+10 \\ &= x^2(y+2) + x(y+2)(6-5y) + (y+2)(6y^2-13y+5) \\ &= (y+2)(x^2+(6-5y)x+(6y^2-13y+5)) \\ &= (y+2)(x-2y +1)(x-3y+5) \end{align*}

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Solution:

Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)

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Solution: \begin{align*} && ( x^2 - a x + b ) ( x^2 + a x + c ) &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && x^4 + p x^2 + q x + r &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && p &= b+c-a^2 \tag{1}\\ && q &= a(b-c) \tag{2}\\ && r &= bc \tag{3} \end{align*} \begin{align*} (1): && p+a^2 &= b+ c \\ (2): && \frac{q}{a} &= b - c \\ \Rightarrow && b &= \frac12 (p+a^2 + \frac{q}{a}) \\ && c &= \frac12 (p+a^2 - \frac{q}{a}) \\ (3): && r &= \frac12 (p+a^2 + \frac{q}{a}) \frac12 (p+a^2 - \frac{q}{a}) \\ \Rightarrow && 4ra^2 &= (pa + a^3 + q)(pa+a^3-q) \\ &&&= (pa+a^3)^2 - q^2 \\ &&&= a^2(p+a^2)^2 -q^2 \\ &&&= a^2(p^2 + 2pa^2 + a^4) - q^2 \\ &&&= pa^2 + 2pa^4 + a^6 - q^2 \\ \end{align*} Therefore \(a^2\) is a root of \(u^3 + 2pu^2 + pu - q^2 = 4ru\), ie the given equation. When \(u = 0\), this equation is \(-q^2\), therefore the cubic is negative. But as \(u \to \infty\) the cubic tends to \(\infty\), therefore it must cross the \(x\)-axis and have a positive root. If \(p=-1, q = -6, r = 15\) then the cubic is: \(u^3 - 2u^2 + (1-60)u -36\) and so when \(u = 9\) we have \begin{align*} 9^3 - 2\cdot 9^2 -59 \cdot 9 -36 &= 9(9^2-2\cdot 9 - 29 -4) \\ &= 9(81 -18-59-4) \\ &= 0 \end{align*} so \(u = 9\) is a root Let \(y=z + 2\) \begin{align*} &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= (z+2)^4-8(z+2)^3 + 23(z+2)^2 - 34(z+2) + 39 \\ &&&= z^4+8z^3+24z^2+32z+16 - \\ &&&\quad -8z^3-48z^2-96z-64 \\ &&&\quad\quad +23z^2+92z+92 \\ &&&\quad\quad -34z-68 + 39 \\ &&&= z^4-z^2-6z+15 \end{align*} So conveniently this is \(p = -1, q = -6, r = 15\), so we know that \(a = 3\) is a sensible thing to true. \(b = \frac12(-1 + 9 + \frac{-6}{3}) = 3\) \(c = \frac12(-1+9-\frac{-6}{3}) = 5\) so \begin{align*} && z^4-z^2-6z+15 &= (z^2-3z+3)(z^2+3z+5) \\ &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= ((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5) \\ &&&= (y^2-4y+4-3y+6+3)(y^2-4y+4+3y-6+5) \\ &&&= (y^2-7y+13)(y^2-y+3) \end{align*}

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Solution: Notice by polynomial division, we can write \(p(x) = (x-a)q(x) + r(x)\) where degree \(r(x) < 1\), ie \(r(x)\) is a constant. Evaluating at \(x = a\), we have \(p(a) = (a-a)q(a) + r(a) = r(a)\). Therefore \(r(a) = p(a)\) and since \(r(x)\) is a constant, it is always \(p(a)\).

1. \(\,\) \begin{align*} && p(x) &= (x-1)(x-2)(x-3)q(x) + r(x) \\ && p(1) &= r(1) = 3 \\ && p(2) &= r(2) = 1 \\ && p(3) &= r(3) = 5 \end{align*} Therefore \(r(x)\) is a polynomial of degree \(2\) or less through \((1,3),(2,1), (3, 5)\) we can write this as \begin{align*} && r(x) &= 5\frac{(x-1)(x-2)}{(3-1)(3-2)} + 1\frac{(x-1)(x-3)}{(2-1)(2-3)} + 3\frac{(x-2)(x-3)}{(1-2)(1-3)} \\ &&&= \frac52(x^2-3x+2)-(x^2-4x+3) + \frac32(x^2-5x+6) \\ &&&= 3x^2-11x+11 \end{align*}
2. Let \(P_n(x) = x(x-1)\cdots(x-n) + x\), then \(P_n(a) = a\)

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Solution: \begin{align*} && p_0(x) &= (1-x)(1-x^2)(1-x^4) \\ &&&= (1-x)(1-x)(1+x)(1-x^2)(1+x^2) \\ &&&= (1-x)^2 (1+x)(1-x)(1+x)(1+x^2) \\ &&&= (1-x)^3 (1+x)^2 (1+x^2) \end{align*} \begin{align*} && p_0'(x) &= 3(1-x)^2(1+x)^2(1+x^2) + (1-x)^3 q(x) \\ \Rightarrow && p_0'(1) &= 3 \cdot 0 \cdots + 0 \cdots \\ &&&= 0 \\ && p_1'(x) &= p_0(x) + xp'_0(x) \\ \Rightarrow && p_1'(1) &= p_0(1) + 1\cdot p_0'(1) \\ &&&= 0 + 1 \cdot 0 \\ &&&= 0 \end{align*} Notice that \(p_0(x) = (1-x-x^2+x^3)(1-x^4) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7\), so: \(p'_0(x) = -1-2x+3x^2-4x^3+5x^4+6x^5-7x^6 \Rightarrow p'_0(1) = 0 = -1 -2 -4 -7 + 3 + 5+6\). \((xp'_1(1))' = 0 = -1^2-2^2-4^2-7^2 + 3^2 + 5^2 + 6^2\). Consider \(q_0(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\), then \((1-x)^4\) is a factor, so in particular we know \(q_0(1), (xq_0(x))'|_{x=1} = 0,(x(xq_0(x))')'|_{x=1} = 0\), and so: \(q_0(x) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7 - x^8+x^9+x^{10}-x^{11}+x^{12}-x^{13}-x^{14}+x^{15}\), and so: \(1^r+2^r+4^r+7^r+8^r+11^r+13^r+14^r = 3^r+5^r+6^r+9^r+10^r+12^r+15^r\) for \(r = 1,2,3\)