Squares of Consecutive Numbers

[jay_shark]

Show that the sum of the squares of 1984 consecutive positive integers cannot be a square .

[jay_shark]

I remember when I first attempted this problem I was completely stuck . This was when I was going into my final year of highschool . Then I remember solving a similar problem that you cannot have the sum of the squares of 13 consecutive positive numbers be a square . I was really pleased when I came up with the following solution .

A perfect square can either have remainders 0 or 1 upon division of 4 . At the time , I wasn't even familiar with mod notation but I knew this fact regarding perfect squares .This means that 12 consecutive squares must have remainder 2 . Therefore the sum of 13 consecutive squares can have only remainders 2 or 3 which is a contradiction .

I tried this similar reasoning with the sum of 1984 consecutive squares and for some reason I couldn't arrive at a contradiction . For several weeks I was stuck until I decided to pick up an introductory number theory book . Then I was trying to find some type of algebraic manipulation and that didn't pan out . Lo and behold , after weeks and months of trying , I finally figured it out .

If you work with mod 16 then the remainders are 0,1,4,9,0,1,4,9 etc . 1984 =16*124 so the sum of the remainders is 8 which is not possible . For the longest time I was losing hope , but I didn't give up on the problem and after countless hours , I finally arrived at a solution .
Ohh the struggles of being a math fanatic .

I hope you enjoyed my story :P

[thylacine]

[ QUOTE ]
I remember when I first attempted this problem I was completely stuck . This was when I was going into my final year of highschool . Then I remember solving a similar problem that you cannot have the sum of the squares of 13 consecutive positive numbers be a square . I was really pleased when I came up with the following solution .


A perfect square can either have remainders 0 or 1 upon division of 4 . At the time , I wasn't even familiar with mod notation but I knew this fact regarding perfect squares .This means that 12 consecutive squares must have remainder 2 . Therefore the sum of 13 consecutive squares can have only remainders 2 or 3 which is a contradiction .

I tried this similar reasoning with the sum of 1984 consecutive squares and for some reason I couldn't arrive at a contradiction . For several weeks I was stuck until I decided to pick up an introductory number theory book . Then I was trying to find some type of algebraic manipulation and that didn't pan out . Lo and behold , after weeks and months of trying , I finally figured it out .

If you work with mod 16 then the remainders are 0,1,4,9,0,1,4,9 etc . 1984 =16*124 so the sum of the remainders is 8 which is not possible . For the longest time I was losing hope , but I didn't give up on the problem and after countless hours , I finally arrived at a solution .
Ohh the struggles of being a math fanatic .

I hope you enjoyed my story :P

[/ QUOTE ]

I don't think this does it, but you are on the right track. The problem is that the sum of the squares of 1984 consecutive positive integers is always a multiple of 16 (do you see why?), so you don't get the desired contradiction.

There do exist positive integers m,n such that the following two statements hold.

(1) The sum of the squares of 1984 consecutive positive integers is always congruent to m mod n.

(2) No perfect square is congruent to m mod n.

See if you can find m and n that work.

[Evenkeal]

lol at this problem

obviously try mod 1984

[jay_shark]

I had the solution mixed up with another problem that required mod 16.

mod 1984 works

0^2+1^2+2^+...1983^3= 1983(1984)*(2*1983+1)/6.

Then use the fact that x^2 cannot equal 992 mod 1984 .

[BruceZ]

[ QUOTE ]
Show that the sum of the squares of 1984 consecutive positive integers cannot be a square .

[/ QUOTE ]

I will prove this for any 1984 consecutive squares, not just the first 1984. The sum of 1984 consecutive squares starting from an arbitrary n^2 is

n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + ... + (n+1983)^2

= 1984*n^2 + 1983*1984 + (1/6)*1983*1984*(2*1983+1)

For the last term, see Power Sum (http://mathworld.wolfram.com/PowerSum.html) for the sum of consecutive squares starting from 1. Now we can remove a factor of 992:

= 992 * [2*n^2 + 2,626,153]

Factoring 992 this becomes

= 2^5 * 31 * [2*n^2 + 2,626,153].

If this were a perfect square, then there would have to be an even number of factors of 2 and 31, and so the term in brackets must contain a factor of 2*31 = 62, but this not possible since the term in brackets is an even number + an odd number = an odd number. Therefore this cannot be a perfect square. QED

[jay_shark]

Thx for your input .

It is also interesting to note that the sum of the squares of 16(2n+1) consecutive positive integers cannot be a perfect square .

The sum of the remainers is 8 upon division by 16 which is a contradiction since x^2 is congruent to 0,1,4,9mod16.

Also , any number of the form 4n+1 cannot be written as the sum of 4n+1 consecutive positive integers .

[BruceZ]

I have an algebra error in my above post, but it doesn't change the nature of the proof. Here is a corrected version, with the change marked in red below:

I will prove this for any 1984 consecutive squares, not just the first 1984. The sum of 1984 consecutive squares starting from an arbitrary n^2 is

n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + ... + (n+1983)^2

= 1984*n^2 + 1983*1984<font color="red">*n</font> + (1/6)*1983*1984*(2*1983+1)

For the last term, see Power Sum (http://mathworld.wolfram.com/PowerSum.html) for the sum of consecutive squares starting from 1. Now we can remove a factor of 992:

= 992 * [2*n^2 + <font color="red">1983*2*n + 2,622,187</font>]

Factoring 992 this becomes

= 2^5 * 31 * [2*n^2 + 1983*2*n + 2,622,187].

If this were a perfect square, then there would have to be an even number of factors of 2 and 31, and so the term in brackets must contain a factor of 2*31 = 62, but this not possible since the term in brackets is <font color="red">even + even + odd</font> = odd. Therefore this cannot be a perfect square. QED

[jay_shark]

I stated that the sum of the squares of 4n+1 consecutive numbers cannot be a perfect square .

This should be 4(2n+1) =8n+4

Using x^2 is congruent 0 or 1 mod 4 .