## imqwerty one year ago fun question (⌐■_■)

1. imqwerty

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

2. anonymous

How is this fun lol

3. imqwerty

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

4. anonymous

Ok we will see about that

5. imqwerty

sure :D

6. anonymous

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

7. 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.

8. anonymous

Are we gonna work it out

9. 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.

10. imqwerty

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

11. dan815

working cases so far

12. dan815

theres no solution for all different

13. 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

14. imqwerty

nope there r 3 sets of solutions :)

15. dan815

7,3,3 1,0,0

16. dan815

oh wait1,0,0 not possible

17. dan815

7,3,3

18. dan815

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

19. imqwerty

no this question is pure algebra no permutations :)

20. dan815

|dw:1439830819873:dw|

21. dan815

well i got that solution from algebra xD

22. dan815

the other solutions have negative numbers in them

23. 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

24. dan815

oh my bad i see l<m<n lol

25. dan815

okay lemme look for solutions in all different case too then

26. 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$

27. dan815

ah i tried somethjing similar with +1

28. dan815

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

29. imqwerty

:D

30. imqwerty

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

31. ganeshie8

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

32. imqwerty

yes @ganeshie8 these r the solutions :)

33. anonymous

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

34. 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.

35. dan815

yes oldrin

36. anonymous

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

37. dan815

oh ok

38. 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*}

39. dan815

|dw:1439833355443:dw|

40. anonymous

41. dan815

ok what now?

42. anonymous

Maybe considerations mod l,m,n

43. dan815

no just algebra its a highschool question

44. anonymous

modular arithmetic is certainly not beyond competition high school math

45. dan815

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

46. imqwerty

yes this question is all algebra :D

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

49. 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

50. 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

51. anonymous

I got lost on the 3rd number o-o

52. dan815

neat trick

53. dan815

with the bounding

54. dan815

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

55. dan815

too much clutter to read through up there

56. 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 :)

57. ali2x2

^ lol gg