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  $P(a)=-P(b)=a-b\;\;or\;\;P(a)=-P(b)=-(a-b)$.

$\blacksquare$