MHB Find Positive Ints Whose Square Ends 444

[kaliprasad]

Find all positive integers whose square end with 444.

 

[kaliprasad]

Find all positive integers whose square end with 444.

Hint

Spoiler

1000 = 8 * 125 and 8 and 125 are co-primes

 

[Albert1]

Find all positive integers whose square end with 444.

Spoiler

there are too many ,I just give 10 examples:
38,462,538,962,1038,1462,1538,1962,2038,2462,------
we can find there exists a rule between those numbers

 

[kaliprasad]

Spoiler

there are too many ,I just give 10 examples:
38,462,538,962,1038,1462,1538,1962,2038,2462,------
we can find there exists a rule between those numbers

Spoiler

Answer is right but 1) closed form of solution to be provided and 2) answer need to be found systematically

 

[Albert1]

Spoiler

there are too many ,I just give 10 examples:
38,462,538,962,1038,1462,1538,1962,2038,2462,------
we can find there exists a rule between those numbers

my solution

Spoiler

first I focus on last two digits 44
since $(40-2)^2=38^2=1444$
and $(60+2)^2=3844$
$(462)^2=(500-38)^2=213444$
so we get two classes
first class :$a_1=38, a_2=538,a_3=1038, a_n=500(n-1)+38\,\,\, (n\in N)$
second class $b_1=(500-a_1)=462,b_2=962,b_3=1462,b_n=500(n-1)+462\,\,\, (n\in N)$

 

[kaliprasad]

my solution

Spoiler

first I focus on last two digits 44
since $(40-2)^2=38^2=1444$
and $(60+2)^2=3844$
$(462)^2=(500-38)^2=213444$
so we get two classes
first class :$a_1=38, a_2=538,a_3=1038, a_n=500(n-1)+38\,\,\, (n\in N)$
second class $b_1=(500-a_1)=462,b_2=962,b_3=1462,b_n=500(n-1)+462\,\,\, (n\in N)$

Spoiler

Answer is right but I am not convinced with the method as this does not conclusively show that above is only solution ( though it is)

here is my solution

Spoiler

we have $x^2 \equiv 444 \pmod {1000}$
as 1000 = 8 * 125 and 8 and 125 are co-primes let us find mod 8 and mod 125
hence $x^2 \equiv 4 \pmod 8\cdots(1) $
hence $x^2 \equiv 69 \pmod {125}\cdots(2) $
from (1) we have $ x \equiv 2 \pmod 4 \cdots(3)$ as x has to be even and x cannot be multiple of 4 then $x^2$ becomes multiple of 16
$x = 4k + 2 => x^2 = 16k^2 + 16 k + 4 => x^2 \equiv 4 \pmod 8$
to solve 2 we get following 2 additional equations
$x^2 \equiv 4 \pmod {5}\cdots(4) $
$x^2 \equiv 19 \pmod {25}\cdots(5) $
to solve (4) we get $x\equiv 2 \pmod {5}\cdots(6) $ and knowing that if x is a solution so is $-x$ by putting 0,1,2
so $x = 5m+2$
so $x^2 = 25m^2 + 20m + 4 \equiv 19 \pmod {25}$
or $20m + 4 \equiv 19 \pmod {25}$
or $20m \equiv 15 \pmod {25}$
or $4m \equiv 3 \pmod {5}$
or $m \equiv 2 \pmod {5}$
so $x \equiv 12 \pmod {25}$
so $x = 25p + 12\cdots(6)$
square to get $x^2 = 625p^2 + 600p + 144 \pmod {125}$
or $100p + 144 \equiv 69 \pmod {125}$
or $100p \equiv 50 \pmod {125}$
this gives $4p \equiv 2 \pmod {5}$ or $p \equiv 3 \pmod {5}$ or = 3
so we get from (6) $x \equiv = 87 \pmod {125} \cdots(7)$
from (3) and (7) we get $x = 125m + 87 \pmod {500}$ and putting m =3 we get $x = 462 \pmod {500}$ or $-38 \pmod {500}$
as specified above $x \equiv 38 \pmod {500}$ ( if x is a solution so is -x)
so solution set $x \equiv \pm 38 \pmod {500}$ and we can chose positive values as 500n+ 38 and 500n + 462 with n zero or positive.

 

[Albert1]

my solution

Spoiler

first I focus on last two digits 44
since $(40-2)^2=38^2=1444$
and $(60+2)^2=3844$
$(462)^2=(500-38)^2=213444$
so we get two classes
first class :$a_1=38, a_2=538,a_3=1038, a_n=500(n-1)+38\,\,\, (n\in N)$
second class $b_1=(500-a_1)=462,b_2=962,b_3=1462,b_n=500(n-1)+462\,\,\, (n\in N)$

Spoiler

let $x^2=?444$
($x^2 $ mod 1000 =444)
the other solution can be obtained by the following procedure:
let $y^2=(500\pm x)^2=250000\pm x\times 1000+x^2)=----444$
($y^2$ mod 1000=444)
and $y$ also will satisfy the need
if one solution is found say $"x=38"$ , all the other solutions can be obtained ,and will be the only solutions
here $x=38$ is easy to get