Question

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?

Answer 1

\[\left(\frac{n(n+1)}{2}\right)^2\]and\[\frac{n(n+1)(2n+1)}{6}\]

Answer 2

\[= \frac{3n(n+1)}{2(2n+1)}\]

Answer 3

oh boy messed up that calculation

Answer 4

i was actually thinking of using quadratic equations for this :)

Answer 5

\(\dfrac{1^3+2^3+...+n^3}{1^2+2^2+...+n^2}=\dfrac{3n(n+1)}{2(2n+1)}\)

i got same calculation @ParthKohli

Answer 6

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

Answer 7

well first condition i could think of is n is even,
so n=2m

Answer 8

oh you number theory fanatics :)

Answer 9

you sure :P it's easiest

Answer 10

but that's not necessary. counterexample n = 1

Answer 11

well for n>1 :)

Answer 12

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 \(n=1\)

Answer 13

nice argument using divisibility rules
but how would one think of doing that

Answer 14

@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

Answer 15

:] 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

Answer 16

interesting, you did long division

Answer 17

yeah :)

Answer 18

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

Answer 19

you can use the same long division argument even if you didn't make this simplification

Answer 20

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 :)

Answer 21

can you show me how you knew to take out n like that?

Answer 22

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-> \(n+\frac{-n^2+n}{4n+2}=n- \frac{n^2-n}{4n+2}\)

Answer 23

:) thats pretty cool

Answer 24

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.

Answer 25

haha thanks (:

Answer 26

but you ended up having to do long division anyway
since that some number , n^2 -n, the divisors are not obvious

Answer 27

yeah

Answer 28

ok it would be like solving
find integer solutions to
12 / ( 4n+2) , so that the fraction is an integer

Answer 29

this technique not exactly this but in one way used in such ques- http://openstudy.com/study#/updates/5604f6eee4b033021d80d8fc

Answer 30

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?)