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Friday, July 14, 2017

Mathematical Reflections 2017, Issue 2 - Problem O406

Problem:
Solve in prime numbers the equation x^3-y^3-z^3+w^3+\dfrac{yz}{2}(2xw+1)^2=2017.

Proposed by Titu Andreescu, University of Texas at Dallas, USA


Solution:
Since x^3-y^3-z^3+w^3 and 2017 are integers, then also \dfrac{yz}{2}(2xw+1)^2 must be an integer. Since (2xw+1)^2 is odd, then 2 \ | \ yz. Since y and z are primes, then y=2 or z=2. By symmetry, we can assume y=2. We have
x^3-z^3+w^3+z(2xw+1)^2=2025. Since 2025 is odd, then the LHS must be odd and it's easy to see that x,z,w cannot be all odd. Moreover, the given equation can be written as x^3+w^3+z(2xw-z+1)(2xw+z+1)=2025. Since z(2xw-z+1)(2xw+z+1) is always even, then x^3+w^3 must be odd, which implies that at least one between x and w is even, i.e. x=2 or w=2. By symmetry, we can assume x=2. We have w^3+z(4w-z+1)(4w+z+1)=2017. If 4w-z+1 \geq 0, then w^3 \leq 2017, so w \leq 11. An easy check gives w=7 and z=2.
So, (x,y,z,w) \in \{(2,2,2,7),(7,2,2,2)\}.
The case 4w-z+1 < 0 remains open.

Mathematical Reflections 2017, Issue 2 - Problem U407

Problem:
Prove that for every \varepsilon>0 \int_2^{2+\varepsilon} e^{2x-x^2} \ dx<\dfrac{\varepsilon}{1+\varepsilon}.

Proposed by Titu Andreescu, University of Texas at Dallas, USA


Solution:
Observe that for all real numbers t \geq 1 we have t^2 \leq e^{t^2-1}. Indeed, put f(t)=e^{t^2-1}-t^2. Then, f(1)=0 and f'(t)=2t(e^{t^2-1}-1) \geq 0 for all t \geq 1. Now, put t=x-1. We have (x-1)^2 \leq e^{x^2-2x} for all x \geq 2, i.e. e^{2x-x^2} \leq \dfrac{1}{(x-1)^2}.
So, \int_2^{2+\varepsilon} e^{2x-x^2} \ dx \leq \int_2^{2+\varepsilon} \dfrac{dx}{(x-1)^2}=\left[-\dfrac{1}{x-1}\right]_2^{2+\varepsilon}=\dfrac{\varepsilon}{1+\varepsilon}.

Mathematical Reflections 2017, Issue 2 - Problem U406

Problem:
Evaluate \lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that \renewcommand{\arraystretch}{2} \begin{array}{rcl} \cos(n+1)x&=&1-\dfrac{1}{2}(n+1)^2x^2+o(x^3) \\  \sin nx&=&nx-\dfrac{1}{6}n^3x^3+o(x^3) \\ n\sin x&=&nx-\dfrac{1}{6}nx^3+o(x^3). \end{array}
So, \dfrac{\left(1-\frac{1}{2}(n+1)^2x^2+o(x^3)\right)\left(nx-\frac{1}{6}n^3x^3+o(x^3)\right)-\left(nx-\frac{1}{6}nx^3+o(x^3)\right)}{x^3}=-\dfrac{\frac{x^3}{3}n(n+1)(2n+1)+o(x^3)}{x^3}. So, \lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}=-\dfrac{n(n+1)(2n+1)}{3}.

Mathematical Reflections 2017, Issue 2 - Problem U403

Problem:
Find all cubic polynomials P(x) \in \mathbb{R}[x] such that P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.

Proposed by Alessandro Ventullo, Milan, Italy


Solution:
Let P(x)=ax^3+bx^2+cx+d be a third-degree polynomial with a,b,c,d \in \mathbb{R}, a \neq 0, such that
P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.
Setting x=-1,x=0,x=1 into the given relation, we obtain
\begin{array}{rcl} P(0)-P^2(-1)+P(1)&=&1 \\ P(1)-P^2(0)+P(-1)&=&1 \\ P(-1)-P^2(1)+P(0)&=&1. \end{array}
Adding these three equations side by side, we get (P(-1)-1)^2+(P(0)-1)^2+(P(1)-1)^2=0, so P(-1)=P(0)=P(1)=1. It follows that
\begin{array}{rcl} d&=&1 \\ a+b+c+d&=&1 \\ -a+b-c+d&=&1. \end{array} We obtain d=1,b=0 and c=-a. So, P(x)=ax^3-ax+1.
Setting x=2 into the given relation, we have P(-6)-P^2(2)+P(4)=1, which gives -36a^2-162a+1=1. Likewise, setting x=-2 into the given relation, we have P(-4)-P^2(-2)+P(6)=1, which gives -36a^2+162a+1=1. Subtracting, we get 264a=0, i.e. a=0. So, we obtain P(x)=1.

Mathematical Reflections 2017, Issue 2 - Problem S407

Problem:
Let f(x)=x^3+x^2-1. Prove that for any positive real numbers a, b, c, d satisfying a+b+c+d>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}, at least one of the numbers af(b), bf(c), cf(d), df(a) is different from 1.

Proposed by Adrian Andreescu, Dallas, Texas, USA


Solution:
Assume by contradiction that af(b)=bf(c)=cf(d)=df(a)=1. The given inequality becomes
\begin{equation}\label{first-ineq} a+b+c+d>f(a)+f(b)+f(c)+f(d). \end{equation}
Observe that f(x) is increasing on (0,+\infty) and f(x)<x if and only if x<1. Moreover, f is convex on (0,+\infty), so by inequality \eqref{first-ineq} and Jensen's Inequality, we have \dfrac{a+b+c+d}{4}>\dfrac{f(a)+f(b)+f(c)+f(d)}{4} \geq f\left(\dfrac{a+b+c+d}{4}\right). It follows that \dfrac{a+b+c+d}{4}<1, i.e. a+b+c+d<4 and from \eqref{first-ineq}, we get
f(a)+f(b)+f(c)+f(d)<4. On the other hand, by the AM-GM Inequality, we have \begin{array}{lll} af(a)+bf(b)+cf(c)+df(d) & \geq & 4 \\ bf(a)+cf(b)+df(c)+af(d) & \geq & 4 \\ cf(a)+df(b)+af(c)+bf(d) & \geq & 4 \\ df(a)+af(b)+bf(c)+cf(d) & \geq & 4. \end{array} Adding these four inequalities, we get (a+b+c+d)(f(a)+f(b)+f(c)+f(d))\geq 16, i.e. f(a)+f(b)+f(c)+f(d) \geq \dfrac{16}{a+b+c+d} > 4, contradiction.
The conclusion follows.

Mathematical Reflections 2017, Issue 2 - Problem S403

Problem:
Find all primes p and q such that \dfrac{2^{p^2-q^2}-1}{pq} is a product of two primes.

Proposed by Adrian Andreescu, Dallas, Texas, USA


Solution:
Since 2^{p^2-q^2}-1 is odd and \dfrac{2^{p^2-q^2}-1}{pq} is an integer, then p and q are odd primes and clearly p>q. So, p^2-q^2 \equiv 0 \pmod{8}. If q>3, then p^2-q^2 \equiv 0 \pmod{3}, so p^2-q^2 is divisible by 24. Then
2^{p^2-q^2}-1=2^{24k}-1, where k \in \mathbb{N}^*. Since 2^{24k}-1 is divisible by 2^{24}-1=16777215=3^2\cdot 5\cdot 7 \cdot 13 \cdot 17 \cdot 241, then \dfrac{2^{p^2-q^2}-1}{pq} cannot be the product of two primes. So, q=3 and
\dfrac{2^{p^2-q^2}-1}{pq}=\dfrac{2^{p^2-9}-1}{3p}. Since p^2-9 is divisible by 8, then p^2-9=8k for some k \in \mathbb{N}^*, so \dfrac{2^{p^2-9}-1}{3p}=\dfrac{2^{8k}-1}{3p}=\dfrac{(2^k-1)(2^k+1)(2^{2k}+1)(2^{4k}+1)}{3p}. An easy check shows that it must be k \geq 2. If k is even, then 3 \ | \ (2^k-1) and if k>2 we have 2^k-1=3n for some n \in \mathbb{N} and n>1. But then
\dfrac{n(2^k+1)(2^{2k}+1)(2^{4k}+1)}{p} is the product of at least three primes, contradiction. So, k=2 and we get p=5, which satisfies the condition. If k is odd, then 3 \ | \ (2^k+1) and since k \geq 3, then 2^k+1=3n for some n \in \mathbb{N} and n>1. But then, we have that \dfrac{n(2^k-1)(2^{2k}+1)(2^{4k}+1)}{p} is the product of at least three primes, contradiction.
In conclusion, (p,q)=(5,3).

Mathematical Reflections 2017, Issue 2 - Problem J408

Problem:
Let a and b be nonnegative real numbers such that a+b=1. Prove that \dfrac{289}{256} \leq (1+a^4)(1+b^4) \leq 2.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam


Solution:
By the Cauchy-Schwarz Inequality, we have (1+a^4)(1+b^4) \geq (1+a^2b^2)^2=(1+a^2(1-a)^2)^2 \geq \left(1+\dfrac{1}{16}\right)^2=\dfrac{289}{256} and the minimum is attained if and only if a=b=\dfrac{1}{2}. By the AM-GM Inequality, we have
(1+a^4)(1+b^4) \leq \dfrac{(1+a^4)^2+(1+b^4)^2}{2}=\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2}.
Using the fact \begin{array}{rcll} a^4 & \leq & (1-a)^4, & a \in [0,1/2] \\ (1-a)^4 & \leq & a^4, & a \in [1/2,1], \end{array} then if a \in [0,1/2], we have
\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+(1-a)^4)^2 \leq 2
and if a \in [1/2,1], we have
\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+a^4)^2 \leq 2.
The conclusion follows.

Mathematical Reflections 2017, Issue 2 - Problem J407

Problem:
Solve in positive real numbers the equation \sqrt{x^4-4x}+\dfrac{1}{x^2}=1.

Proposed by Titu Andreescu, University of Texas at Dallas, USA


First solution:
The equation has a real solution if and only if x \leq -1 or x \geq \sqrt[3]{4}. We have x^4-4x=\left(1-\dfrac{1}{x^2}\right)^2 \iff x^4-4x+\dfrac{4}{x^2}=\left(1-\dfrac{1}{x^2}\right)^2+\dfrac{4}{x^2}, i.e.
\left(x^2-\dfrac{2}{x}\right)^2=\left(1+\dfrac{1}{x}\right)^2
So, x^2-\dfrac{2}{x}=\pm \left(1+\dfrac{1}{x^2}\right).
Observe that x^2-\dfrac{2}{x}=-1-\dfrac{1}{x^2} gives x^4+(x-1)^2=0, which has no real solutions. So,
x^2-\dfrac{2}{x}=1+\dfrac{1}{x^2} \iff x(x^3-1)=x^2+x+1 \iff x(x-1)=1. So, we get x^2-x-1=0, which yields x=\dfrac{1+\sqrt{5}}{2}.

Second solution:
The given equation is equivalent to x\sqrt{x^4-4x}=x-\dfrac{1}{x}. Squaring both sides, we get x^6-4x^3=x^2-2+\dfrac{1}{x^2}. Hence, (x^3-2)^2=\left(x+\dfrac{1}{x}\right)^2. Since x^3>4, we have x^3-2=x+\dfrac{1}{x}. So, x^4-2x=x^2+1, which gives x^4=(x+1)^2, i.e. x^2=x+1. So, x=\dfrac{1+\sqrt{5}}{2}.

Mathematical Reflections 2017, Issue 2 - Problem J406

Problem:
Let a,b,c be positive real numbers such that a+b+c=3. Prove that a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution:
Let f(x)=x\sqrt{x+3}. Observe that f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0. Therefore, by Jensen's Inequality, we have 2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3}, i.e. a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.