Question

Find the first integer \(n\ge 1\) such that the average
of \(1^2;2^2;3^2...n^2\) is itself a perfect square.

Answer 1

Keep in mind that the "average" of a set of numbers is \[average=\frac{ \sum~of~the~numbers }{ count~of~the~numbers }\]

Answer 2

for example, the average of 1^2, 2^2 and 3^2 is\[average = \frac{ 1^2+2^2+3^2 }{ 3 }=\frac{ 1+4+9 }{ 3 }=\frac{ 14 }{ 3 }\] Is (14/3) a perfect square? No? Then you'll have to keep going; next would be the average of the squares of 1, 2, 3 and 4. I don't know of a more sophisticated approach; perhaps someone else does.

Nevertheless, by continuing to find the averages as explained here, you'll likely eventually find the desired integer n.

Answer 3

the trivial answer is n = 1

for other solutions, we may try to mess wid :

\(\large \frac{(n+1)(2n+1)}{6} = k^2\)

\(\large (n+1)(2n+1) = 6k^2\)

Answer 4

i feel stuck... @KingGeorge

Answer 5

Hey, ganeshie8, that's a great approach!! I'd suggest you look up "summation formula n^2" or something equivalent. There is definitely a formula for the sum of the terms k^2, but I've forgotten it.

Once we have that formula, it'd be an easy matter to let n = 2, 3, 4, ..., until we find a sum that is itself a perfect square. More power to you!

Answer 6

See http://polysum.tripod.com/

You'll find the necessary summation formula for k^2 there.

Answer 7

yeah looks we're not going to get lucky wid small numbers

the first solution after n = 1 is n= 337 which is a very messy prime lol

Answer 8

How'd you arrive at n=337?

Answer 9

For the lack of a more sophisticated approach, I'd set up an Excel spreadsheet showing the sums of k^2 for n=1, 2, 3, etc. Grunt work. But it'd surely show us quickly for which n the sum of k^2 would itself be a perfect square.

Answer 10

can we use some elementary number theory to find the roots,like this \(6|(n+1)(2n+1)\)

Answer 11

trial and error

Answer 12

lets see...

2n + 1 is odd
so 2 | (n+1)

Answer 13

which implies that n is itself odd,otherwise that will not hold

Answer 14

yahhh thats pretty much it, we cannot figure out anything further from that i think...

Answer 15

hmmm.....ok ,as alternative,is it not enough for us to seek to write \(\dfrac{(n+1)(2n+1)}{6}\) in the form \((an+b)^2\) for \(a,b\) real numbers

Answer 16

this doesnt seem to be possible :/

Answer 17

another try at this....it seems like if we try to solve this using quadratic equation we have

\[n=\dfrac{-3+\sqrt{1+48k^2}}{2(2)}\implies \]
\[1+48k^2\] hould also be a square

Answer 18

(4n+3)^2 = 48k^2 + 1

Answer 19

n=1,k=1
so we also remember that n is odd so
\[4n+3=4(2j-1)+3=(8j-1)\]


\[(8j-1)^2=48k^2+1\implies 8j(8j-2)=48k^2\]

\[j(4j-1)=3k^2\]

Answer 20

is this taking us to square zero,i mean back to our original problem? or leading somewher

Answer 21

i'm taking it to SE,i will update it as it gets solutions there :)

Answer 22

Ohhk... good luck :) plz provide the SE link also.....

Answer 23

http://math.stackexchange.com/questions/751968/find-the-first-integer-n-ge-1-such-that-the-average-of-12-22-32-n2-is

Answer 24

thanks a lot ,its nice brainstorming with you,u have great approach to problem solving Sir

Answer 25

lol dont make me Sir xo
Jonask ever tried quadratic diaphantine equations ?

Answer 26

yes i have tried them hmmmm ok now i have a new approach all together,but will need some help

Answer 27

i only studied diophantine of the form \[ax^2+bxy+cy^2=m\]

Answer 28

I have been lazy to complete the last 4 chapters in NT.. which cover all these cool stuff :/

looks the equation u got wid quadratic formula can be thought of as a quadratic :

\((8j-1)^2=48k^2+1\)

\((8j-1)^2-48k^2+1\)

Answer 29

*\((8j-1)^2-48k^2 = 1 \)

Answer 30

i think it should be easy for u to solve the quadratic diaphantine....
for me it wud take some time to udnerstand the mechanics of quadratic equations....

Answer 31

this is perfect,now we have found an organized solution,actually after u sujjested this,i think i dont need SE

Answer 32

http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html

x^2 - Dy^2 = 1

Answer 33

i will take some time as well,i'm sure wolfram can do a better job

Answer 34

wow ! plz show the solution once u have it ! would love to see how to get n=337 analytically xD