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lfdahl
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Find all natural numbers $n$, for which:
\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]
\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]
my solution:lfdahl said:Find all natural numbers $n$, for which:
\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]
lfdahl said:Find all natural numbers $n$, for which:
\[\sqrt{1 \underbrace{4...4}_{n \: times}} \in \Bbb{N}\]
kaliprasad said: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
Albert said:$10000n$ shouled be replaced with $10^n$ how do you think ?
kaliprasad said: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
Thanks a lot, Albert. I am sure, yours and kaliprasad´s solution are correct. Please allow me to askAlbert said:my solution:
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 !)
This is also my questionlfdahl said:Thankyou very much, kaliprasad. I am quite sure, that your solution is correct. Please allow me to
ask two questions:
(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 said:Thanks a lot, Albert. I am sure, yours and kaliprasad´s solution are correct. Please allow me to ask
a question:
In case b = -2, you conclude, that the leftmost digit of a is 4. Why?
lfdahl said:Thankyou very much, kaliprasad. I am quite sure, that your solution is correct. Please allow me to
ask two questions:
(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")?
Albert said: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 said: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.
Natural numbers are positive integers (whole numbers) starting from 1 and increasing with no upper limit. They do not include fractions or decimals.
To find all natural numbers, you can start from 1 and continue counting up, or use a mathematical formula such as the set notation N = {1, 2, 3, ...} where the three dots indicate that the numbers continue infinitely.
Natural numbers include all positive integers starting from 1, while whole numbers also include 0. In other words, all natural numbers are whole numbers, but not all whole numbers are natural numbers.
Yes, natural numbers and counting numbers refer to the same set of numbers. Both include all positive integers starting from 1.
Finding all natural numbers is important in various mathematical concepts and calculations, such as in number patterns, prime factorization, and solving equations. It also helps in understanding the concept of infinity and the concept of one-to-one correspondence in counting.