Question

Find any 3 consecutive integers such that each of them is divisible by a perfect square > 1

Answer 1

For example : first integer can be divisible by 4, second integer by 25 and third integer by 49

Answer 2

hmm,if we have an integer like \(t\) then \( t \equiv 0,1,2,3 ~ (mod ~ 4) \) so \( t^2 \equiv 0,1 \) but as we have 3 consecutive numbers none of the numbers can be divisible by the same numbers : suppose \( s^2 | a \) and \( s^2 | a+2\) so \( s^2 | 2 \) and it can never happen.

Answer 3

therefore we have 3 perfect square which all of them are different and we have 2 cases for being congruent (mod 4) , using pigeonhole principle we can say two of the square numbers are congruent (mod 4)

Answer 4

well,maybe this can be used as the idea ;D work on it.

Answer 5

yes! we can't say that two perfect squares are divisble by 4 because then we can say 4|a and 4|a+1 or a+2 and it can't happen so, \( t^2 \equiv s^2 \equiv 1 ~ (mod ~ 4 ~ )\)

Answer 6

thats right! the divisors must be distinct as the same number cannot divide a, a+1 and a+2

Answer 7

well @ganeshie8 i'm not sure this can be a good idea or not!

Answer 8

I am thinking something like this
suppose 4 divides a
9 divides a+1
25 divides a+2

Answer 9

u only need one set of numbers?

Answer 10

yes just looking for one set of solution :)

Answer 11

\(\Large \begin{align} \color{black}{
548 ~mod ~2^2 \equiv 0\hspace{.33em}\\~\\
549~ mod ~3^2 \equiv 0\hspace{.33em}\\~\\
550 ~mod ~5^2 \equiv 0\hspace{.33em}\\~\\
}\end{align}\)

Answer 12

correct ;) how did u find?

Answer 13

thats very fast ! xD

Answer 14

yes , actually their can be many of examples of these

this is an example of chinese remainder theorm

yesterday @ganeshie8 taught me a problem these so i had the feeling

Answer 15

oh! yes chinese remiander thm but @mathmath333 that thm only say that there most be an integer which is divisble by n perfect square.

Answer 16

it won't give u the numbers...

Answer 17

* must

Answer 18

well u could consider \(n^2\) as any integers ,ex,4 9 ,36, 49,64,81

Answer 19

another example

\(\Large \begin{align} \color{black}{
15849 ~mod ~3^2 \equiv 0\hspace{.33em}\\~\\
15850 ~mod ~5^2 \equiv 0\hspace{.33em}\\~\\
15851~ mod ~11^2 \equiv 0\hspace{.33em}\\~\\
}\end{align}\)

Answer 20

Looks neat :)
here is one way to work it with out using chinese remainder thm :
\[\begin{align} a&\equiv 0 \pmod 4\\~\\
a+1&\equiv 0 \pmod{9}\\~\\
a+2&\equiv 0 \pmod{25}

\end{align}\]

from the first congruence we have \(a = 4m\)
plugging that in second congruence we get \(4m + 1\equiv 0 \pmod{9}\)
solving gives \(m = 2 + 9n\)

plugging this back into \(a\) we get \(a = 4(2+9n)\)
we can plug this into third congruence and solve...

Answer 21

yes^^

Answer 22

@ganeshie8 thanks ;)
@mathmath333 nice way ;)

Answer 23

i think the value of the three squares should be coprime or should be squares of prime numbers ,is that correct @ganeshie8

Answer 24

thats right! to apply chinese remainder theorem we need the moduli to be corprime to each other

Answer 25

yes so something like \(2^2,4^2,16^2\) wont work here

Answer 26

\(a\equiv 0 \pmod {2^2} \)
\(a+1\equiv 0 \pmod {4^2} \)

itself is not solvabel because the gcd of 2^2 and 4^2 is 2^2
and 2^2 does not divide -1

Answer 27

we have this rule :
\[x\equiv a \pmod n\\x\equiv b \pmod m\]

this system is solvable only when \(\gcd(m,n)\) divides \(a-b\)

Answer 28

I'll post a separate question for its proof
it is fun :)

Answer 29

from the first equation we have \( x = nk + a \) and from the second \( x = mk^* + b\) so \(a-b + nk - mk^*= 0 \) therefore \( a-b = mk^* - nk\) but \( gcd(m,n) | mk^* \) and \(gcd(m,n) | nk \) so \( a-b|gcd (m,n) \)

Answer 30

but besides this we have to prove that when \( a-b | gcd(m,n)\) then that system is solvable.

Answer 31

then we can say that system is resolvable when and only when \( a-b|gcd(m,n) \)

Answer 32

thats exactly same as the proof im thinking of ! xD

\[a-b = mk^* - nk\]

suppose gcd(m,n) = d, then we can factor this out from the right hand side :
\[a-b = d(stuff)\]


clearly \(a-b\) is a multiple of \(d\)
which is same as saying \(d\) divides \(a-b\)

Answer 33

i'm thinking for the second part of the proof.suppose a-b|d then prove that system is resolvable.

Answer 34

Ahh yes that looks bit involved

Answer 35

yes,maybe doing the same thing as we do for proving chineese remiander thm!

Answer 36

@PFEH.1999

\(\Large \begin{align} \color{black}{
a~ mod ~2^2 \equiv 0\hspace{.33em}\\~\\
a+1 ~mod ~3^2 \equiv 0\hspace{.33em}\\~\\
a+2 ~mod ~5^2 \equiv 0\hspace{.33em}\\~\\
}\end{align}\)

this is same as

\(\Large \begin{align} \color{black}{
a ~mod ~2^2 \equiv 0\hspace{.33em}\\~\\
a ~mod ~3^2 \equiv 8\hspace{.33em}\\~\\
a~ mod ~5^2 \equiv 23\hspace{.33em}\\~\\
}\end{align}\)


thats why we apply chinese rem thrm

Answer 37

and from this we can find a ?

Answer 38

yes of cos

Answer 39

yes thank toy ;) beautiful way ;)

Answer 40

oh lol sry it \(\huge 548\) please check it

Answer 41

correct

Answer 42

thnks

Answer 43

and would u write ur complete solution i don't know why i get wrong answer!

Answer 44

yes here is a problem like that and their are 3 to 4 methods for solving it,
have a look
http://openstudy.com/study#/updates/54aabff5e4b0b83d9dca0684

Answer 45

thanks ;)

Answer 46

thinking of Quadratic residue

Answer 47

\(n|p_1^2\\n-1|p_2^2\\n+1|p_3^2\\n(n-1)(n+1)|(p_1p_2p_3)^2~~~~~\text{s.t p1,p2,p3 can be different or the same }\)

Answer 48

@mathmath333 , a=4k , 4k = 8 ( mod 9) so k=2 (mod 9) so k=9s + 2 and a would be 36s + 8 thus 25 | 36s - 15 so 25|12s - 5

Answer 49

how to solve s and plug in to find a?

Answer 50

\(\large \begin{align} \color{black}{
a ~mod ~2^2 \equiv 0\hspace{.33em}\\~\\
a ~mod ~3^2 \equiv 8\hspace{.33em}\\~\\
a ~mod~ 5^2 \equiv 23\hspace{.33em}\\~\\

\normalsize \text{by the chinese remainder theorm}\hspace{.33em}\\~\\

a=x+y+z~(mod~m)\hspace{.33em}\\~\\
x=0\cdot 9\cdot 25\left( (9\cdot 25)^{-1}~(mod~4) \right)\hspace{.33em}\\~\\
x=0\hspace{.33em}\\~\\
y=8\cdot 4\cdot 25\left( (4\cdot 25)^{-1}~(mod~9) \right)\hspace{.33em}\\~\\
y=8\cdot 4\cdot 25\left(1 \right)\hspace{.33em}\\~\\
z=23\cdot 4\cdot 9\left( (4\cdot 9)^{-1}~(mod~25) \right)\hspace{.33em}\\~\\
z=23\cdot 4\cdot 9\left( 16 \right)\hspace{.33em}\\~\\
m= 4\cdot 9\cdot 25\hspace{.33em}\\~\\
a=\left(0+8\cdot 4\cdot 25\left(1 \right)+23\cdot 4\cdot 9\left( 16 \right)\right) ~(mod~4\cdot 9\cdot 25)\hspace{.33em}\\~\\
a=\dfrac{8\cdot 4\cdot 25 }{4\cdot 9\cdot 25}+\dfrac{23\cdot 4\cdot 9\left( 16 \right) }{4\cdot 9\cdot 25}\hspace{.33em}\\~\\
a=\dfrac{8 }{ 9}+\dfrac{23\left( 16 \right) }{ 25}\hspace{.33em}\\~\\
a=\dfrac{-1 }{ 9}+\dfrac{-7}{ 25}\hspace{.33em}\\~\\
a=\dfrac{-88 }{ 9\cdot 25}\\~\\
a=\dfrac{-88\cdot 4 }{ 9\cdot 25\cdot 4 }\\~\\
a=\dfrac{-352}{ 900 }\\~\\
a=\dfrac{900-352}{ 900 }\\~\\
a=\dfrac{548}{ 900 }\\~\\
a=548
}\end{align}\)