ogeometrie

vineri, 2 decembrie 2022

Problem 28423 from GMB 10/2022

The problem is proposed for the 9th grade, on page 483.

In translation, thanks to Miss Google:

"Show that the equation $x^{n+1}+y^n=z^n$ has an infinity of solutions in the set of nonzero natural numbers, for any natural number $n \geqslant 2$."

ANSWER CiP

$$x=\frac{(k^{n+1}+1)^n-1}{k^n},\;\;y=\frac{(k^{n+1}+1)^n-1}{k^{n+1}},\;\;z=\frac{(k^{n+1}+1)^{n+1}-k^{n+1}-1}{k^{n+1}}$$

represent solutions of the given equation, for $k \in \mathbb{N}\setminus \{0\}.$

Solution CiP

Let us first show that the values given in the Answer are solutions of the equation.

Let $l \underset{def}{=} k^{n+1}+1.$ Then $x=\frac{l^n-1}{k^n},\;y=\frac{l^n-1}{k^{n+1}},\;z=\frac{l^{n+1}-l}{k^{n+1}}$ and we have

$$x^{n+1}+y^n=\frac{(l^n-1)^{n+1}}{k^{n(n+1)}}+\frac{(l^n-1)^n}{k^{(n+1)n}}=$$

$=\frac{(l^n-1)^n}{k^{(n+1)n}}\cdot [(l^n-1)+1]=\frac{(l^n-1)^n \cdot l^n}{k^{(n+1)n}}=\left ( \frac{l(l^n-1)}{k^{n+1}} \right )^n=\left ( \frac{l^{n+1}-l}{k^{n+1}} \right ) ^n\;=\;\;z^n.$

We found the expressions in the answer by trying to find solutions of the given equation with $x=ky,\;k \in \mathbb{N} \setminus \{ 0 \}.$ We have

$$x^{n+1}+y^n=k^{n+1} \cdot y^{n+1}+y^n=y^n(k^{n+1} \cdot y+1).$$

For the above result to be a power of $n$ we must have $k^{n+1}y+1=l^n$ for some integer $l$, so $y=\frac{l^n-1}{k^{n+1}}.$ For $y$ to be an integer, it is enough to choose $l=k^{n+1}+1$ because according to the binomial theorem we have

$l^n-1=(k^{n+1}+1)^n-1=\sum_{i=0}^n \textrm{C}_n^i (k^{n+1})^{n-i}=$

$=[(k^{n+1})^n+\textrm{C}_n^1(k^{n+1})^{n-1}+ \cdots +\textrm{C}_n^{n-1}k^{n+1}+1]-1=k^{n+1} \cdot A$,

for some integer $A$. That's how I found the answer.

$\blacksquare$

REMARK CiP

In particular cases, the solutions can be expressed differently. For $n=2$ the equation

$$x^3+y^2=z^2$$

$\Leftrightarrow \;x^3=(z-y)(z+y)$ can be solved by choosin

$$\begin{cases}z-y=x\\z+y=x^2\end{cases}$$

obtaining an infinite family of complete solutions $$x=m,\;y=\frac{m^2-m}{2},\;z=\frac{m^2+m}{2},\;m\in \mathbb{Z}.$$

(I did not investigate if these represent the complete solutions of the equation.)

The particular case has the following solutions

The general formula provides the solutions in the case of n=2

miercuri, 30 noiembrie 2022

Problem E:16386 from GMB 10/2022

The problem proposed for class 6, on page 480.

In translation (thanks to Miss Google):

"Let $p$ and $q$ be two distinct prime numbers. Prove that if $15(p+q) >8pq$, then

$$93pq \leqslant 174(p+q) \leqslant 145 pq."$$

ANSWER CiP

One of the numbers is $2$

and the other can be $3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29.$

SOLUTION CiP

Natural numbers that verify the condition

$$15(p+q)>8pq \tag{1}$$

must be in finite quantity. Indeed, the condition (1) can be written equivalent to

$$\frac{1}{p}+\frac{1}{q}>\frac{8}{15}. \tag{2}$$

If none of the numbers $p,q$, is equal to $2$, then $\frac{1}{p}+\frac{1}{q} \leqslant \frac{1}{3}+\frac{1}{5}=\frac{8}{15},$ contradicting (2).

Assuming $p=2$ the condition (1) is written $15(2+q)>16q\;\;\Leftrightarrow\;\;q<30$. So $3\leqslant q \leqslant 29$ and $q$ is a prime number. Then we have

$$\frac{1}{2}+\frac{1}{29}\leqslant \frac{1}{p}+\frac{1}{q}\leqslant \frac{1}{2}+\frac{1}{3}\;\Leftrightarrow\;\frac{31}{58}\leqslant \frac{p+q}{pq}\leqslant \frac{5}{6}$$

and multiplying the last inequalities by $6 \cdot 29=174$ we get the conclusion.

$\blacksquare$

REMARK CiP

The condition (1) can also be written, after multiplying by 8, $$64pq-120p-120q<0\;\Leftrightarrow$$

$\Leftrightarrow \;(8p-15)(8q-15)<225$. If $p\geqslant 3$, we have $8p-15 \geqslant 9$ then we deduce $8q-15<\frac{225}{8p-15}\leqslant \frac{225}{9}=25$, so $8q<40$, $q<5$. We only have the possibility $q=3$ or $q=2$. When $q=3$, from (1) we deduce $p<5$. We can only have $p=3$ but then $p=q$, impossible. When $q=2$, from (1) we deduce $p<30$ and we find the answer.

$\square$ end REM