fun question (⌐■_■)

- imqwerty

fun question (⌐■_■)

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- schrodinger

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- imqwerty

Determine all triplets (l,m,n) of positive integers such that \[l \le m \le n\]and\[l+m+n+lm+mn+nl = lmn+1\]

- anonymous

How is this fun lol

- imqwerty

:D right nw this problem looks neat but nasty but once u see the solution u'll smile :)

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## More answers

- anonymous

Ok we will see about that

- imqwerty

sure :D

- anonymous

How is it that you got so many Medals and we ain't done nothing yet

- Loser66

@isaac4321 since it is interesting problem. Students learn a lot from it. The medal is for the Askers who have a good question also.

- anonymous

Are we gonna work it out

- Loser66

Thanks @imqwerty your problems inspired me a lot. They tell me how not good I am, how stupid I am so that I have to try harder on my studying.

- imqwerty

welcome @Loser66 i'll be posting such problems time to time :) nd hey ur smart :D

- dan815

working cases so far

- dan815

theres no solution for all different

- dan815

looks like theres solutions for 2 same and 1 different, i think theres an argument here for all different not being possible something intuitively tells me that

- imqwerty

nope there r 3 sets of solutions :)

- dan815

7,3,3
1,0,0

- dan815

oh wait1,0,0 not possible

- dan815

7,3,3

- dan815

and its permutations?
3,7,3
3,3,7

- imqwerty

no this question is pure algebra no permutations :)

- dan815

|dw:1439830819873:dw|

- dan815

well i got that solution from algebra xD

- dan815

the other solutions have negative numbers in them

- dan815

and i think theres a AM GM argument here about why all different is not possible for hte positive integers
so the solution has to exist for
2 same, 1 different case only
(L,m,m) once u find a solution for this u can multiple it by 3 since its
m,L,m
m,m,L same thing
so we have to find integer solutions for this equation
L+2m+2*mL +m^2=m^2L+1

- dan815

oh my bad i see l

- dan815

okay lemme look for solutions in all different case too then

- imqwerty

a hint -
u can start it like this :) -
\[l-1=p\]\[m-1=q\]\[n-1=r\] where \[p,q,r \in integers \]\[0 \le p \le q \le r\]

- dan815

ah i tried somethjing similar with +1

- dan815

but got lazy and said i hope theres just no solution for all different

- imqwerty

:D

- imqwerty

the start which i gave is jst a partial start :) u need to find a way to get through

- ganeshie8

if it helps, there are only 3 solutions
|dw:1439831929718:dw|

- imqwerty

yes @ganeshie8 these r the solutions :)

- anonymous

$$(1 + l)(1 + m)(1 + n) = 1 + l + m + n + lm + ln + mn + lmn$$
so this is equivalent to $$(1+l)(1+m)(1+n)=2(1 + lmn)$$

- thomas5267

I noticed that in each of the solution, two pairs are coprime and one isn't. For example, 2 and 8 are not coprime and (2,5) and (5,8) are both coprime.

- dan815

yes oldrin

- anonymous

I wasn't asking for confirmation, I'm suggesting for an actual approach

- dan815

oh ok

- thomas5267

Is the following true?
\[
\begin{align*}
l+m+n+lm+mn+nl &= lmn+1\\
l+m+n+lm+mn+nl - lmn&=1\\
\implies \gcd(l,m,n,lm,mn,nl,lmn)=1\\
\end{align*}
\]
\[
\begin{align*}
\gcd(l,m,n,lm,mn,nl,lmn)&=1\\
\gcd(\gcd(l,lm,nl,lmn),\gcd(m,mn),n)&=1\\
\implies \gcd(l,m,n)&=1
\end{align*}
\]

- dan815

|dw:1439833355443:dw|

- anonymous

yes, those follow readily from the thing I wrote above

- dan815

ok what now?

- anonymous

Maybe considerations mod l,m,n

- dan815

no just algebra its a highschool question

- anonymous

modular arithmetic is certainly not beyond competition high school math

- dan815

xD im just saying for this question its pretty much all algebra

- imqwerty

yes this question is all algebra :D

- anonymous

I can't really type well on my phone but I think it helps if you write
$$(1-l)(1-m)(1-n)=1-(l+m+n)+lm+ln+mn-lmn\\(l-1)(m-1)(n-1)=-1+l+m+n-(lm+ln+mn)+lmn\\(l-1)(m-1)(n-1)=(lmn+1)-(l+m+n+lm+ln+nm)+2(lmn+l+m+n)\\(l-1)(m-1)(n-1)=2(lmn+l+m+n)$$
Or something along those lines

- anonymous

Oops , it should read \(2(1+l+m+n)=2(4+p+q+r)\) so$$pqr=2(4+p+q+r)$$

- anonymous

I probably made a few typos typing on my phone since OpenStudy lags so much and I can't even see what I'm typing but something like thet

- imqwerty

\[l-1=p\]\[m-1=q\]\[n-1=r\]so the equation can be written in the form\[pqr=2(p+q+r)+4\]
where p,q,r are integers such that\[0 \le p \le q \le r\]. we can clearly see that p=0 is not possible so we can write the equation-\[pqr=2(p+q+r)+4\] as\[2(1\div pq + 1\div qr + 1\div rp) + 4\div pqr = 1\].
IF \[p \ge 3\] then \[q \ge 3 \]and\[r \ge 3\]
so the left side is bounded by 6/9+4/27 which is less than 1. We conclude that p=1 or 2.
CASE 1:-
suppose p=1 then we have \[qr=2(q+r)+6\]or \[(q-r)(r-2)=10\]This gives \[q-2=1 \],\[r-2=10\] OR \[q-2=\],\[r-2=5\] recall\[q \le r\]So this implies (p,q,r)=(1,3,12) or (1,4,7).
CASE 2-
if p=2 the equation reduces to (q-1)(r-1)=5
hence q-1=1 and r-1=5 is the only solution. this gives (p,q,r)=(2,2,6)
reverting back to l,m,n finall :D we get 3 triplets - (2,4,13), (2,5,8) and (3,3,7)
:D

- anonymous

I got lost on the 3rd number o-o

- dan815

neat trick

- dan815

with the bounding

- dan815

can u show me how u got to this equation again
pqr=2(p+q+r)+4

- dan815

too much clutter to read through up there

- imqwerty

i just substituted the values of l as p+1 nd similarly m nd n nd then solved the equation nd finally the result is
pqr=2(p+r+r)+4 :)

- ali2x2

^ lol gg

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