How come sum of 2 consecutive whole sqs can be a wholesq, but not so with 3,4,5,6,7,8,9 or 10 consecutive sqs?

Question

We know 3^2 + 4^2 = 5^2
and also 20^2 + 21^2 = 29^2
Such cases not there in sum of more than 2 consecutive wholesqs.
However, sum of 11 consecutive whole sqs can be a whole sq itself
e.g. 18^2 + 19^2 ....+28^2 = 77^2

Answer 1

Well, the case of 3 is easy to show. Suppose n is an integer such that n^2 + (n+1)^2 + (n+2)^2 is a square.

Expanding those terms, you get:

3n^2 + 6n + 5

Which is:

3(n^2+2n+2) - 1

But there are no squares of the form 3K-1.

In general, the sum of m consecutive squares can be written as:

m * n^2 + m(m-1) * n + m(m-1)(2m-1)/6

Setting this equation equal to K^2, then multiplying both sized by 36 and dividing by m, we get the equation:

36n^2 + 36(m-1)n + 6m(m-1)(2m-1) = 36K^2/m

Complete the square on the left side:

(6n+3(m-1))^2 +6(m-1)(2m-1)-9(m-1)^2 = 36K^2/m

(6n+3(m-1))^2 +3(m^2-1) = 36K^2/m

Dividing by 9 and multiplying by m we:

m(2n+(m-1))^2 + m(m^2-1)/3 = (2K)^2

Let V=2n+(m-1) and U=2K. Then we can write this equation:

U^2-mV^2 = m(m^2-1)/3

Case m=4:

U^2-4V^2 = 20, or

(U-2V)(U+2V) = 20

So we'd need to factor 20 into two numbers that differ by a multiple of four, which is not possible.

m=5:

U^2 - 5 V^2 = 5*4*2 = 40

So U is divisble by 5, and we can let U=5W and re-arrange the equation:

V^2 = 5W^2 -8 = 5(W^2-2) + 2

But there are no squares of the form 5Z+2.

m=6:

U^2 - 6V^2 = 70

Looking mod 7 that means U^2 = -V^2 (mod 7), but -1 is not a square mod 7. No solutions.

m=7:

U^2 - 7V^2 = 7*8*2 = 112

so U must be divisible by 7, let U=7W, and we can rewrite this as:

V^2 = 7W^2 -16 = 7 (W^2-3) + 5

But there are no squares of the form 7Z+5.

m=8:

U^2 - 8V^2 = 8*7*3 = 168

U must be divisible by 4, U=4W.

V^2 -2W^2 = -21

Mod 7, that means V^2 = 2W^2. But 2 is not a square mod 7.

m=9:

U^2 - 9V^2 = 3*80

So U must be divisible by 3, and then the LHS is divisible by 9, but the RHS is not divisible by 9.

m=10:

U^2 -10V^2 = 10*11*3

U^2 + V^2 = 0 (mod 11)

but -1 is not a square mod 11, so this has no solutions.

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Oh, and here's a solution for the sum of 11 squares:

18^2 + 19^2 + ... + 28^2 = 77^2

The next case for 11 starts with 456.

For most m's, if there is a solution, there are infinitely many solutions. For m a perfect square, however, there always at most finitely many solutions.

Answer 2

In the case of 3, you can show three consecutive number with (a-1), a, (a+1) if a≥2. Thus, sum of these 3 consecutive whole sqs would be:
(a-1)² + a² + (a+1)² = a² - 2a + 1 + a² + a² + 2a + 1=3a² + 2
If you imagine that 3a² + 2 is a whole-sq, then you can write:

3a² + 2 = b² → a² = (b²-2)/3
It means that b²-2 should be dividable to 3. Do you think it's possible? In other words, is there any number like b that b²-2 could be dividable to 3? let solve this problem.
The number of "b" can be in three different position:
1- It could be dividable to 3 so b=3n. Thus, b²-2 = 9n²-2 = 3(3n²) - 2. It means b²-2 is not dividable to 3. So b can't be 3n.
2- It could be 1 more than a number which is dividable to 3 so b=3n+1. Therefore, b²-2 = 9n²+6n-1 = 3(3n²+2n) - 1. It means b²-2 is not dividable to 3. So b can't be 3n+1.
3- It could be 2 more than a number which is dividable to 3 so b=3n+2. so b²-2 = 9n²+12n+2 = 3(3n²+4n) + 2 and its not dividable to 3 neither.

Finally you cam get the result that b can't be equal to 3n, 3n+1 or 3n+2. In other world there is not any number like b that 3a² + 2 = b² and it shows to us that there are not 3 consecutive numbers that sum of their whole Sqs be a whole sq number.

Answer 3

102 and 104 are the numbers. you are able to write the smaller selection as n. simply by fact the bigger selection is a consecutive even integer, it extremely is written as n+2. once you upload them at the same time, you get 206, it is represented by: (n) + (n+2) = 206 2n + 2 = 206 resolve for n 2n = 204 n = 102 considering n = 102, then n+2 = 104. those are your 2 numbers.