A RATIONAL VALUE of COSINE // مقدار گویای کسینوس

               In the magazine GAZETA MATEMATICĂ No 4, 1978, on covers 3-4 the following Problem appears (as a solution to the one proposed in GMB 11/1977, page 455 Problem#3) :

          Problem 3 page 455 / No. 11/1977  : Prove that if   $\alpha \in \mathbb{R}$  and 

      $\cos(\alpha \pi)=\frac{1}{3}$  then  $\alpha$  is irrational. (The angle  $\alpha \pi$ is considered in radians)

(We will return to this issue at the end)

               In the past, in my youth, I was more diligent. In the magazine GMB 10/2014 (so 35 years after the one mentioned in the preamble), in the -recently established- Column PROBLEMS for NATIONAL EXAMS, in the 12th grade, page 478, we see problems 27 and 28. (We made a system of cards with them, by grade, etc.; that's why I said "diligence")

This statement

Solving

As you can see, in Problem 2c), the question arises whether at the value  $q\pi$  for which  $\cos q\pi=\frac{3}{5}$, we have  $q\in\mathbb{Q}$  or not. When resolving this point, I remembered the problem in the Preamble. The answer is NO.

So, $\cos q\pi=\frac{3}{5}\;\Rightarrow\; q\notin \mathbb{Q}$.

         We assume the opposite (see line (1) in the third picture) : 

$\exists q\in \mathbb{Q}\;\;s.t.\;\;\cos q\pi=\frac{3}{5}\tag{1}$

Let  $q=\frac{m}{n},\;m\in\mathbb{Z},\;n\in \mathbb{N}$, with their greatest common divisor  $(m,n)=1$. Then

 the set of values ​​of the string  $\{\cos kq\pi;\;q\in\mathbb{Z}\}$ is finite :                           (2)

$\pm\cos 0q\pi,\;\pm\cos q\pi,\;\pm\cos 2q\pi,\dots,\;\pm\cos (n-1)q\pi \tag{3}$

     Indeed, we will show that, whatever  $k\in\mathbb{Z},\;\cos kq\pi$  has one of the values  (3). Writing, with Euclid's division lemma,  $k=n\cdot t+l,\;\;t\in\mathbb{Z},\;l\in\{o,\;1,\;\dots,\;n-1\}$,  we have

$\cos kq\pi=\cos k(\frac{m}{n})\pi=\cos (nt+l)\frac{m}{n}\pi=\cos (mt+l\frac{m}{n})\pi=\cos (mt\pi+lq\pi)=\pm\cos lqt$

qed$\square$

       On the other hand, we will show that if  $k$ is a power of 2, then  $\cos k\pi$  takes on an infinity of values. More precisely, we show, by mathematical induction, the equality :

$\cos 2^kq\pi=\frac{a_k}{5^{2^k}},\;\;\;a_k\in\mathbb{Z},\;\;\;5\not{\mid}a_k \tag{4}$

For $k=0,\;\;\cos 2^0q\pi=\cos q\pi=\frac{3}{5}=\frac{3}{5^{2^0}}$, and we have  $a_0=3,\;5\not{\mid }a_0.$ Then  $\cos 2q\pi=2\cos ^2q\pi-1=2(\frac{3}{5})^2-1=-\frac{7}{25}=\frac{-7}{5^{2^1}}$, so  $a_1=-7\;and\;5\not{\mid}a_1$.

Assuming (4) true for a fixed  $k$, we have

$\cos 2^{k+1}q\pi=\cos 2(2^kq\pi)=2\cos^2(2^kq\pi)-1=2\cdot (\frac{a_k}{5^{2^k}})^2-1=\frac{2a_k^2-5^{2^k\cdot 2}}{5^{2^k\cdot 2}}=\frac{2a_k^2-5^{2^{k+1}}}{5^{2^{k+1}}}$

and taking  $a_{k+1}=2a_k^2-5^{2^{k+1}}$  we have  $5\not{\mid}a_{k+1}$  because  $5\not{\mid}a_k$. So (4) is thrue for  $k+1$. Moreover, the values ​​given by (4) are all distinct, because if we had  $\frac{a_k}{5^{2^k}}=\frac{a_l}{5^{2^l}},\;k<l$, it would result  $a_l=5^{2^l-2^k}\cdot a_k$  so  $5\mid a_l$ - false.

$\blacksquare$

               The same argument is used in the original problem in the Preamble, where, from  $\cos \alpha \pi=\frac{1}{3}$,  results :

$\cos 2^k\alpha \pi=\frac{r}{3^{2^k}},\;\;r\in\mathbb{Z},\;\;\;3\not{\mid}r$

Problem 2954 from REVISTA MATEMATICĂ a ELEVILOR din TIMIȘOARA

 You can see issue 1-1977 from which I took the issue (page 62), here.

A larger collection is here.

             2 954.  Determine the real numbers  $x$  for which the number

$\frac{x}{x^2-5x+7}$  is an integer.

(Matematika v Șkole)

ANSWER CiP

$x\in \{0,\;2,\;\frac{7}{3},\;3,\;\frac{7}{2}\}$

                    Solution CiP

          If  $\frac{x}{x^2-5x+7}=k\in\mathbb{Z}$  then  $x$  is the root of the quadratic equation

$kx^2-(5k+1)x+7k=0 \tag{1}$

The equation (1) has the discriminant  $\Delta_x=(5k+1)^2-28k^2=1+10k-3k^2$.

     But  $1+10k-3k^2=\frac{28}{3}-\left (\frac{25}{3}-10k+3k^2\right )=\frac{28}{3}-3\cdot \left(\frac{25}{9}-\frac{10}{3}k+k^2\right)=$

$=\frac{28}{3}-3\cdot \left (\frac{5}{3}-k\right)^2\leqslant \frac{28}{3}$,  so  

$\Delta_x\leqslant \frac{28}{3}\tag{2}$

     For the equation (1) to have rational roots, it must be  $\sqrt{\Delta_x}\in\mathbb{Q}$,  in fact  $\sqrt{\Delta_x}\in\mathbb{N}$. This happens if  (cf. (2))

$\sqrt{\Delta_x}\in\{0,\;1,\;2,\;3\} \tag{3}$

     Solving the equations  $1+10k-3k^2=0\;or\;1\;or\;2\;or\;3$  we obtain the integer values

$k\in\{0,\;2,\;3\}$

for which, we obtain from the equation (1) the values ​​of  $x$  in the answer.

$\blacksquare$

          Correction

          My answer is incomplete, as the commenter noted. 

          For  $x=3\pm\sqrt{2}$, the number  $\frac{x}{x^2-5x+7}$  takes the value  $1$, also an integer.

I wrongly assumed that  $x$  is a rational number. Let's analyze the sign of  $\Delta_x$.

          Setting the condition  $\Delta_x \geqslant 0\;\Rightarrow\; 1+10k-3k^2\geqslant 0$  which gives us the admissible values

$k\in\left ( \frac{5-\sqrt{28}}{3}\;,\;\frac{5+\sqrt{28}}{3}\right )\cap \mathbb{Z}=\{0,\;\color{Red}1,\;2,\;3\}$

Considering that the roots of the equation (1) are  $x_{1,2}=\frac{5k+1\pm\sqrt{\Delta_x}}{2k}$, for  $k=1$  we find  $x=3\pm\sqrt{2}$, so the correct answer is

$x\in \{0,\;3-\sqrt{2},\;2,\;\frac{7}{3},\;3,\;\frac{7}{2},\;3+\sqrt{2}\}$

$\color{Red}{!!!}\square \color{Red}{!!!}$

PROBLEM E : 6185 from the magazine GAZETA MATEMATICĂ

          From GMB 4/1978 page 164.

In translation :

                         "E : 6185*.  Knowing that the numbers  4830,  448230,  44482230

         are products of two consecutive natural numbers, find these consecutive

                      numbers. Generalization.

[Author : ] I. I. Mihailov, Ivanovo, U.R.S.S."

ANSWER CiP

$4\;830\;=\;69\;\times\;70$

$448\;230\;=\;669\;\times\;670$

$44\;482\;230\;=\;6\;669\;\times\;6\;670$

[CiP addition] $4\;444\;822\;230\;=\;66\;669\;\times\;66\;670$

                                  Generalization :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\;8\;\underset{n}{\underbrace{2\;\dots\;2}}\;30\;=\;\underset{n+1}{\underbrace{6\;\dots\;6}}\;9\times \underset{n}{\underbrace{6\;\dots\;6}}\;70 \tag{G}$

                         Solution CiP

           If a number  $A$  is a product of consecutive numbers, i.e.  $A=k(k+1)$, then we have

$A=k(k+1) \Rightarrow k^2<A<(k+1)^2\Rightarrow k<\sqrt{A}<k+1\Rightarrow k=[\sqrt{A}]$

([X]=the integer part of the number X) We have :

$\sqrt{4830}=69,4...$  and it is verified by calculation that  $4830=69\times 70$

$\sqrt{448230}=669,4...$  and it is verified by calculation that  $448230=669\times 670$

$\sqrt{44482230}=6669,4...$  and it is verified by calculation that  $44482230=6669\times 6670$

The next example in the answer, verified by calculation, is my creation. I also formulated the generalization, which we will verify below.

          To carry out the calculations we need to replace  $1\;\dots\;1$  with the more precise expression

$\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot (10^n-1) \tag{1}$

Indeed  $\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot \underset{n}{\underbrace{9\;\dots\;9}}=\frac{1}{9}\cdot (1\underset{n}{\underbrace{0\;\dots\;0}}-1)=\frac{1}{9}\cdot(10^n-1).$

          Analyzing the position of the digits of the number  $4\dots482\dots230$, we obtain that the value of the number on the left of the relation  (G) is :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\underset{n+3}{\underbrace{\;8\;\overset{n+2}{\overbrace{\underset{n}{\underbrace{2\;\dots\;2}}\;\underset{2}{\underbrace{30}}}}}}=\underset{n+1}{\underbrace{4\dots4}}\cdot 10^{n+3}+8\cdot 10^{n+2}+\underset{n}{\underbrace{2\dots2}}\cdot 100+30=$

$\overset{(1)}{=}4\cdot \frac{1}{9}\cdot (10^{n+1}-1)\cdot 10^{n+3}+8\cdot 10^{n+2}+2\cdot \frac{1}{9}\cdot(10^n-1)\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}-\frac{4}{9}\cdot 10^{n+3}+8\cdot 10^{n+2}+\frac{2}{9}\cdot 10^{n+2}-\frac{2}{9}\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{2}$

     On the right side of (G) we have :

$\underset{n+1}{\underbrace{6\dots6}}9\times \underset{n}{\underbrace{6\dots6}}70=\left [\frac{6}{9}\cdot (10^{n+1}-1)\cdot 10+9\right ]\times \left[\frac{6}{9}\cdot (10^n-1)\cdot 100+70\right ]=$

$=\left [\frac{2}{3}\cdot 10^{n+2}+\frac{7}{3}\right]\cdot \left [\frac{2}{3}\cdot 10^{n+2}+\frac{10}{3}\right ]=\frac{4}{9}\cdot 10^{2n+4}+\frac{2}{3}\cdot 10^{n+2}\cdot \left (\frac{7}{3}+\frac{10}{3}\right )+\frac{70}{9}=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{3}$

The results in (1) and (2) prove the equality in (G).

$\blacksquare$

A PROBLEM that JUMPED like a KANGAROO

           The author of the problem is George STOICA from Canada.

Although proposed for 12th grade (the last grade before college), it only requires elementary knowledge of quadratic equations and some common sense knowledge of natural numbers.

          Here we denote  $\mathbb{N}=\{0,\;1,\;2,\dots \}$  be the set of natural numbers,  $\mathbb{Q}$  the set of rational numbers. We state the following, almost obvious, statement as a

              Lemma.   Let  $m\in\mathbb{N}$  such that  $\sqrt{m}\in\mathbb{Q}$.    Then  $\sqrt{m}\in\mathbb{N}$.

                  Proof CiP  $\sqrt{m}$ admits the representation  $\sqrt{m}=\frac{q}{r},\;\;q,\;r\in\mathbb{N},\;r\neq0$, where the greatest common divisor of the numbers  $q$ and  $r$,  $(q,r)=1$. By the Fundamental Theorem of Arithmetic, we have the (unique) writings  

$m=p_1^{e_1}\cdot p_2^{e_2}\cdot \dots \cdot p_k^{e_k},\;\;q=p_1^{f_1}\cdot p_2^{f_2}\cdot \dots \cdot p_k^{f_k},\;\;r=p_1^{g_1}\cdot p_2^{g_2}\cdot \dots \cdot p_k^{g_k}\;\;\;\;k\geqslant 1 \tag{1}$

where  $p_1<p_2<\dots <p_k$ are primes and  $e_{1-k},\;f_{1-k},\;g_{1-k}$  are nonnegative integers. In (1), some  $e_i,\;f_i,\;g_i$  can be zero and the condition  $(q,r)=1$  requires that, for any index  $i$, one of  $f_i\; or\; g_i$  be zero.

From the equality  $r^2\cdot m=q^2$  and the uniqueness of the representation as a product of powers of prime numbers, it follows

$2\cdot g_i+e_i=2\cdot f_i$

so all the exponents  $e_i$  are even numbers and then  $\sqrt{m}=p_1^{e_1/2}\cdot p_2^{e_2/2}\cdot \dots \cdot p_k^{e_k/2} \in\mathbb{N}$.

$\square$Lemma

             The issue originally appeared in the GMB Supplement 3/2025, as S:L25.144 on page 15. Then he jumped (like a kangaroo !!) in GMB 6-7-8/2025 , as 29 181 on page 374. (Nice jump, otherwise the problem would not have benefited from ever publishing a solution.) The problem statement is :

                    "Let  $P$  be a polynomial with integer coefficients and  $a,\;b\in\mathbb{Z}$  such 

               that  $P(a)\cdot P(b)=-(a-b)^2$. Show that  $P(a)=-P(b)=\pm(a-b)$."

ANSWER CiP

An EXAMPLE of such a polynomial is  $P(X)=2\cdot X-a-b$

                        Solution (as it appears in GMB 1/2026, page 38, adapted by CiP)

               If  $a=b$, that is,  $P(a)^2=0$, we have  $P(a)=0$  so the conclusion is obvious.  Next we assume  $a\neq b$.

                From the algebraic identity

  $\frac{a^k-b^k}{a-b}=a^{k-1}+a^{k-2}b+\dots+ab^{k-2}+b^{k-1},\;\;\forall k\geqslant 1$

it follows that for any polynomial  $P$ , the number  $\alpha=\frac{P(a)-P(b)}{a-b}\in \mathbb{Z}$.

          For rational numbers 

$x_1=\frac{P(a}{a-b},\;\;\;x_2=-\frac{P(b}{a-b} \tag{2}$

we have  $x_1+x_2=\frac{P(a)-P(b)}{a-b}=\alpha,\;\;x_1\cdot x_2=-\frac{P(a)\cdot P(b)}{(a-b)^2}=1$ so the numbers (2) are the roots of the equation with integer coefficients

$x^2-\alpha \cdot x+1=0. \tag{3}$

Then the discriminant of the equation (3) is a rational number. But  $\Delta=\alpha^2-4$  and then

$\mathbb{Q}\ni \sqrt{\alpha^2-4}=\beta\;\;\overset{\textbf{Lemma}}{\Rightarrow}\;\; \beta \in \mathbb{N}$

We have   $\alpha^2-\beta^2=4=(\alpha+\beta)(\alpha-\beta)$, so,  $\alpha+\beta\;and\;\alpha-\beta$  having the same parity, it results

$\alpha+\beta=\pm2=\alpha-\beta$

and from here  $\beta=0,\;\alpha=\pm2$.

          So from the equation (3) we get  $x_1=x_2=1\;or\;x_1=x_2=-1$. Therefore

$1\overset{(2)}{=}\frac{P(a)}{a-b}=-\frac{P(b)}{a-b}$

or

$-1\underset{(2)}{=}\frac{P(a)}{a-b}=-\frac{P(b)}{a-b}.$

It results from here  $P(a)=-P(b)=a-b\;\;or\;\;P(a)=-P(b)=-(a-b)$.

$\blacksquare$