Andrika's Conjecture versus Rodika's Conjecture : On squares ending with three of the digit 4

           Andrica's conjecture is not a joke. Rodica's conjecture is a joke.

          We read in an article from the mathematics magazine "Revista Matematica a Elevilor din Timisoara", issue 1 of 1975: 

             "It is observed that the squares of some natural numbers end with identical digits (e.g. $12^2=144,\;1038^2=1\;077\;444$). 

     In the same issue of the magazine Dorin ANDRICA, then a diligent student from Deva, published some proposed problems, on pages 41-49: #2102, #2116, #2124, #2134, #2149.

     The cited article "Asupra unei proprietăți a numerelor naturale"(On a Property of Natural Numbers) with author N. I. NEDIȚĂ, on pages 3-6, explains the phenomenon. If a perfect square ends with dentical digits, these digits can only be 44 or 444. The general form of these numbers is, in the first case

$$50*k-38,\;\quad50*k-12,\quad \quad k=1,\;2,\;\dots \tag{1}$$

and in the second case

$$500*k-462,\;\quad500*k-38,\quad \quad k=1,\;2,\;\dots \tag{2}$$

There are no perfect squares that end with the group of digits 4444.

           Remark CiP  Formula (2) is included in Formula (1). To be more precise, certain values ​​of $k$ must be removed from (1).

 

          RODICA, an imaginary character, reads this article and makes the following Conjecture:

          "If a perfect square ends with three digits equal to 4, then the thousands digit of this square can only be odd."

     See the examples: $38^2=\color{Red} 1\;444,\;462^2=21\color{Red}3\;444,\;962^2=92\color{Red}5\;444$

$1038^2=1\;07\color{Red}7\;444,\;538^2=28\color{Red}9\;444$

'Neața NEAȚĂ ! (meaning "Good morning...")

          Ion NEAȚĂ from Slatina is also the author of Problem S:E24.271, published in the same issue of GMB, but this time in the Exercise Supplement (on page 7, proposed for 8th grade).

In translation:

  "S:E24.271  Knowing that $x,\;y\;$ and $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11}$ 

      are natural numbers, show that A is prime."

Answer CiP

$$x=1,\;\;y=3,\;\;\;A=11$$

Solution CiP

(obtained from a collaborator on AOPS)

          We first show the following:

              Lemma If  $a\;$ and $\;b\;$ are natural numbers s.t. $\sqrt{a}+\sqrt{b} \in \mathbb{N}\;$

                            then $a$ and $b$ are perfect squares.

               Proof (CiP) It is known that "the square root of $X$ is rational if and only if X is a rational number that can be represented as a ratio of two perfect squares" (see Wikipedia).

Now, if $\sqrt{a}+\sqrt{b}=m\in \mathbb{N}$ then $\sqrt{b}=m-\sqrt{a}$ and squaring it gives
$$\sqrt{a}=\frac{m^2+a-b}{2m}\in \mathbb{Q}$$

(the excluded case $m=0$ is trivialy). Hence $a$ is perfect square; but then $\sqrt{b}=m-\sqrt{a} \in \mathbb{Q}$ and so $b$ is a perfect square too.

$\square$ Lemma

          According to the Lemma, $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11} \in \mathbb{N}\;\Rightarrow$

$4x^2+2y+6=$ perfect square

$4y^2+2x+11=$ perfect square

so we have the inequalities (the first perfect square being an even number)

$$4x^2+2y+6 \geqslant (2x+2)^2,\;\;\;4y^2+2x+11 \geqslant (2y+1)^2$$

Adding the two inequalities

$$(4x^2+2y+6)+(4y^2+2x+11)\geqslant (4x^2+8x+4)+(4y^2+4y+1)\;\;\Leftrightarrow$$

$$\Leftrightarrow\;12\geqslant 6x+2y\;\;\;\Rightarrow \;\;\;\;y\leqslant 3(2-x). $$

Obvious $x \ngtr 2$. If $x=2\;\Rightarrow\;y=0$, but $4x^2+2y+6=22\neq$perfect square.

If $x=1\;\Rightarrow\;y\leqslant 3\;$ and $\;4x^2+2y+6=2y+10\leqslant 16$ it is a perfect square only for $y=3$.

If $x=0\;\Rightarrow\;y\leqslant 6\;$and $\;4x^2+2y+6=2y+6\leqslant 18$; it is possible only $y=5$ but then $4y^2+2x+11=100+11=11\neq$perfect square.

We got the answer.

$\blacksquare$

If we write QED it means it is GEOMETRY ??

 QED = GEOMETRY

More and more people are hesitant to use the expression QED at the end of a demonstration.

          Regarding problem E:17017 from GMB 10/2024.

In translation:

               "E:17017. Let $ABC$ be any triangle with $AB=3 \cdot BC$. Consider the 

 points $D\in AC$ such that $AD= 3 \cdot DC$, and $E$ the midpoint of the side $AB$.

 Show that $BD \perp DE$."

Solution CiP

          Due to the dimensions given in the problem it is convenient to choose $DC=p$ and $AB=6\cdot m$, hence $AD=3 \cdot p,\;BE=AE=3\cdot m,\;BC=2 \cdot m$. 

          It is obvious that $\frac{AE}{AC'}=\frac{3m}{4m}=\frac{3}{4}=\frac{3p}{4p}=\frac{AD}{AC}$ so with the Reciprocal of Thales' Theorem

$$DE \parallel CC' \tag{1}$$

          It is also obvious that $\frac{BC}{BA}=\frac{2m}{6m}=\frac{1}{3}=\frac{p}{3p}=\frac{CD}{AD}$ so with the Converse of Angle Bisector Theorem $BD$ is the bisector of angle $\measuredangle CBA$.

Then, in the obvious isosceles triangle $BCC'$ the bisector of angle at vertex $B$ is also the altitude, so $BD \perp CC'$. From (1) it then follows $BD \perp DE$.

QED

$\blacksquare$

            Remark CiP

          A more computational solution uses the Law of Cosines.

     In $\Delta ABC\;: \quad cos\; C=\frac{CB^2+CA^2-AB^2}{2\cdot CB \cdot CA}=\frac{4m^2+16p^2-36m^2}{2\cdot 2m\cdot 4p}=\frac{p^2-2m^2}{mp}$

and in $\Delta BCD\;:\quad BD^2=CB^2+CD^2-2\cdot CB \cdot CD \cdot cos\;C=$

$=4m^2+p^2-2 \cdot 2m \cdot p \cdot \frac{p^2-2m^2}{mp}=4m^2+p^2-4(p^2-2m^2)$ so

$$BD^2=12m^2-3p^2 \tag {2}$$

Again in $\Delta ABC\;:\quad cos\;A=\frac{AB^2+AC^2-BC^2}{2\cdot AB \cdot AC}=\frac{36m^2+16p^2-4m^2}{2\cdot 6m \cdot 4p}=\frac{p^2+2m^2}{3mp}$

and in $\Delta ADE\;: \quad DE^2=AD^2+AE^2-2\cdot AD \cdot  AE \cdot cos\;A=$

$=9p^2+9m^2-2\cdot 3p \cdot 3m \cdot \frac{p^2+2m^2}{3mp}=9p^2+9m^2-6(p^2+2m^2)$ so

$$DE^2=3p^2-3m^2 \tag{3}$$

From (2) and (3) it result

$$BD^2+DE^2=(12m^2-3p^2)+(3p^2-3m^2)=9m^2=BE^2$$

hence by Converse of Pythagorean Theorem the angle between sides $BD$ and $DE$ is a right angle.

<end Rem>