anonymous
  • anonymous
For how many natural numbers \(n\) the expression\[\dfrac{1^3+2^3+...+n^3}{1^2+2^2+...+n^2}\]becomes a natural number?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ParthKohli
  • ParthKohli
\[\left(\frac{n(n+1)}{2}\right)^2\]and\[\frac{n(n+1)(2n+1)}{6}\]
ParthKohli
  • ParthKohli
\[= \frac{3n(n+1)}{2(2n+1)}\]
ParthKohli
  • ParthKohli
oh boy messed up that calculation

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ParthKohli
  • ParthKohli
i was actually thinking of using quadratic equations for this :)
ikram002p
  • ikram002p
\(\dfrac{1^3+2^3+...+n^3}{1^2+2^2+...+n^2}=\dfrac{3n(n+1)}{2(2n+1)}\) i got same calculation @ParthKohli
ParthKohli
  • ParthKohli
ok now i'm a little unwell... consider this: take that expression as \(y\) now express it as a quadratic in \(n\) then find condition for integral roots which is an expression in \(y\) which gives you a condition
ikram002p
  • ikram002p
well first condition i could think of is n is even, so n=2m
ParthKohli
  • ParthKohli
oh you number theory fanatics :)
ikram002p
  • ikram002p
you sure :P it's easiest
ParthKohli
  • ParthKohli
but that's not necessary. counterexample n = 1
ikram002p
  • ikram002p
well for n>1 :)
anonymous
  • anonymous
another way to do this\[4n+2 | 3n^2+3n \\ 4n+2|4(3n^2+3n)-3n(4n+2)=6n \\ 4n+2|3(4n+2)-2(6n)=6 \\ 4n+2|6\]only possibility is \(n=1\)
anonymous
  • anonymous
nice argument using divisibility rules but how would one think of doing that
anonymous
  • anonymous
@ParthKohli Can you explain your solution a bit more, how would you take a quadratic in terms of y. I am asking because I see a rational function
bubblegum.
  • bubblegum.
:] a similar way \[n-\frac{ \left( n^2-n \right) }{ 4n+2 }\] dividing \(n^2-n\) by \((4n+2)\) must give a number less than n but doing the division we get quotient=\(\frac{n}{4}-\frac{3}{8}\) remainder=\(\frac{3}{4}\) so this can never be possible but it can be when n=1 cuz then \(n^2-n\) will be 0
anonymous
  • anonymous
interesting, you did long division
bubblegum.
  • bubblegum.
yeah :)
anonymous
  • anonymous
that is nice you simplified the expression $$\dfrac{3n(n+1)}{2(2n+1)} = n-\frac{ \left( n^2-n \right) }{ 4n+2 }$$ what made you think of this
anonymous
  • anonymous
you can use the same long division argument even if you didn't make this simplification
bubblegum.
  • bubblegum.
well doing that won't make much difference but whenever i see such problems my 1st step is to change the format soo i jst did that here too :)
anonymous
  • anonymous
can you show me how you knew to take out n like that?
bubblegum.
  • bubblegum.
okay so my aim was to do something of this format-> \(n + \frac{some~number}{4n+2}\) and if we could do this then we can directly check out the divisors of that "\(some~number\)" and then put them equal to \((4n+2)\) and get possible values of n so my 1st step was to check out what this value is-> \(n+\frac{1}{4n+2}\) simplifying we get->\(\frac{4n^2+2n}{4n+2}+\frac{1}{4n+2}\) to make it equal to the original expression we must replace that 1 by \((-n^2+n)\) so just replace it and we get this-> \(n+\frac{-n^2+n}{4n+2}=n- \frac{n^2-n}{4n+2}\)
anonymous
  • anonymous
:) thats pretty cool
bubblegum.
  • bubblegum.
but here we notice that, that "\(some~number\) is in terms of n and has no common factors to cancel with \(4n+2\) so i did division.
bubblegum.
  • bubblegum.
haha thanks (:
anonymous
  • anonymous
but you ended up having to do long division anyway since that some number , n^2 -n, the divisors are not obvious
bubblegum.
  • bubblegum.
yeah
anonymous
  • anonymous
ok it would be like solving find integer solutions to 12 / ( 4n+2) , so that the fraction is an integer
bubblegum.
  • bubblegum.
this technique not exactly this but in one way used in such ques- http://openstudy.com/study#/updates/5604f6eee4b033021d80d8fc
anonymous
  • anonymous
From parth approach\[\frac{3n(n+1)}{2(2n+1)}=m\]rearrange to get a quadratic in \(n\)\[3n^2+(3-4m)n-2m=0\]For getting integer values for \(n\) discriminant of quadratic must be a complete square\[\Delta_n=16m^2+9=k^2\]only possibility is \(m=1\) (why?)

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