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. $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$