MHB Natural Numbers with Repeating Digits: Solving for Perfect Squares

[lfdahl]

Find all natural numbers $n$, for which:

\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]

 

[Albert1]

Find all natural numbers $n$, for which:

\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]

my solution:

Spoiler

let $(a\pm b)^2=y^2=\[{1 \underbrace{4...4}_{n \: times}} \]$ is a perfect square
here $b=2$
we have :$y^2-4=a(a\pm 4)$
if $a(a+4)=y^2-4=\[{1 \underbrace{4...4}_{m \: times}}0 $
then the leftmost digit of $a=1$
if $a(a-4)=y^2-4=\[{1 \underbrace{4...4}_{m \: times}}0 $
then the leftmost digit of $a=4$
(for the leftmost digit of $y^2-4 = 1$)
as we can see the only possible values for $a=10,or \, 40$
and the solutions of $m=1,or \,2$
$m=1,140=10\times 14$ and $ n=2$ we have $144=12^2$
$m=2,1440=40\times 36$ and $n=3$ we have $1444=38^2$
if $n\geq 4 $ we have $y^2=10^n+4\times\[{ \underbrace{1...1}_{n \: times}} \]$
=$4\times \,\, (3\, mod \,4)$
which can not be a perfect square
for $ (3\, mod \,4)$ is not a perfect square
**(I take a reference of kaliprasad' proof for the last part many thanks !)

 

[kaliprasad]

Find all natural numbers $n$, for which:

\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]

Spoiler

there cannot be more than 3 4;s because 10000n + 4444 = 4 (2500n + 1111)

2500n + 1111 mod 4 = 3 and a square cannot be be 3 mod n

so we are left with 1,2 or 3 4's

14 is not a pwefect square
$144 = 12^3$
$1444 = 38^2$

so we get the numbers 144 or 1444

 

[Albert1]

Spoiler

there cannot be more than 3 4;s because 10000n + 4444 = 4 (2500n + 1111)

2500n + 1111 mod 4 = 3 and a square cannot be be 3 mod n

so we are left with 1,2 or 3 4's

14 is not a pwefect square
$144 = 12^3$
$1444 = 38^2$

so we get the numbers 144 or 1444

Spoiler

$10000n$ shouled be replaced with $10^n$ how do you think ?

 

[kaliprasad]

Spoiler

$10000n$ shouled be replaced with $10^n$ how do you think ?

No

Spoiler

I have taken 10000n + 4444 and not n 4's) n 4's is a special case of hat

.

 

[lfdahl]

Spoiler

there cannot be more than 3 4;s because 10000n + 4444 = 4 (2500n + 1111)

2500n + 1111 mod 4 = 3 and a square cannot be be 3 mod n

so we are left with 1,2 or 3 4's

14 is not a pwefect square
$144 = 12^3$
$1444 = 38^2$

so we get the numbers 144 or 1444

Thankyou very much, kaliprasad. I am quite sure, that your solution is correct. Please allow me to
ask two questions:

Spoiler

(1). You say: "A square cannot be 3 mod n", I would replace "n" by "4". Am I right??
(2). What is meant by the number: 10000n+4444? Is it a multiplication: 10000xn? If yes, then why is this used and not:
10^n + 4 x 11..1 (n times "1")?

 

[lfdahl]

my solution:

Spoiler

let $(a\pm b)^2=y^2=\[{1 \underbrace{4...4}_{n \: times}} \]$ is a perfect square
here $b=2$
we have :$y^2-4=a(a\pm 4)$
if $a(a+4)=y^2-4=\[{1 \underbrace{4...4}_{m \: times}}0 $
then the leftmost digit of $a=1$
if $a(a-4)=y^2-4=\[{1 \underbrace{4...4}_{m \: times}}0 $
then the leftmost digit of $a=4$
(for the leftmost digit of $y^2-4 = 1$)
as we can see the only possible values for $a=10,or \, 40$
and the solutions of $m=1,or \,2$
$m=1,140=10\times 14$ and $ n=2$ we have $144=12^2$
$m=2,1440=40\times 36$ and $n=3$ we have $1444=38^2$
if $n\geq 4 $ we have $y^2=10^n+4\times\[{ \underbrace{1...1}_{n \: times}} \]$
=$4\times \,\, (3\, mod \,4)$
which can not be a perfect square
for $ (3\, mod \,4)$ is not a perfect square
**(I take a reference of kaliprasad' proof for the last part many thanks !)

Thanks a lot, Albert. I am sure, yours and kaliprasad´s solution are correct. Please allow me to ask
a question:

Spoiler

In case b = -2, you conclude, that the leftmost digit of a is 4. Why?

 

[Albert1]

Thankyou very much, kaliprasad. I am quite sure, that your solution is correct. Please allow me to
ask two questions:

Spoiler

(1). You say: "A square cannot be 3 mod n", I would replace "n" by "4". Am I right??
(2). What is meant by the number: 10000n+4444? Is it a multiplication: 10000xn? If yes, then why is this used and not:
10^n + 4 x 11..1 (n times "1")?

This is also my question

 

[Albert1]

Thanks a lot, Albert. I am sure, yours and kaliprasad´s solution are correct. Please allow me to ask
a question:

Spoiler

In case b = -2, you conclude, that the leftmost digit of a is 4. Why?

Spoiler

if $b=-2,$ we have :
$(a+b)^2=a^2-4a+4,y^2-4=a^2-4a=a(a-4)$
for the leftmost digit of $a(a-4)$ is 1
so the leftmost digit of $a$ must be 4
and the leftmost digit of $(a-4)$ must be 3
$ a >0,and \, (a-4)>0$

 

[kaliprasad]

Thankyou very much, kaliprasad. I am quite sure, that your solution is correct. Please allow me to
ask two questions:

Spoiler

(1). You say: "A square cannot be 3 mod n", I would replace "n" by "4". Am I right??
(2). What is meant by the number: 10000n+4444? Is it a multiplication: 10000xn? If yes, then why is this used and not:
10^n + 4 x 11..1 (n times "1")?

Spoiler

1) you are right it is a typo

2) there are 2 ways to look at it

I say that it cannot end with 4444( this is 4 4's)say we have 144444 = 14 * 10000 + 4444 = 4*( 14 * 2500 + 1111)

now 14 * 2500 + 1111 = 3 mod 4 and cannot be a perfect squares

this includes all the numbers that you have specified and also 134444 which is additional your number specified and not left out any of your numbers

hence my approach is also correct.

 

[lfdahl]

Spoiler

if $b=-2,$ we have :
$(a+b)^2=a^2-4a+4,y^2-4=a^2-4a=a(a-4)$
for the leftmost digit of $a(a-4)$ is 1
so the leftmost digit of $a$ must be 4
and the leftmost digit of $(a-4)$ must be 3
$ a >0,and \, (a-4)>0$

Thankyou for the answer, Albert! - and for your participation. Good job!

 

[lfdahl]

Spoiler

1) you are right it is a typo

2) there are 2 ways to look at it

I say that it cannot end with 4444( this is 4 4's)say we have 144444 = 14 * 10000 + 4444 = 4*( 14 * 2500 + 1111)

now 14 * 2500 + 1111 = 3 mod 4 and cannot be a perfect squares

this includes all the numbers that you have specified and also 134444 which is additional your number specified and not left out any of your numbers

hence my approach is also correct.

Thankyou for the answer, kaliprasad! - and for your participation. Good job too!