Can one split the 81 first squares into 9 groups of 9 squares each with identical sum?

Well here goes all the sets as required.

1st: 1^2 + 2^2 + 3^2 + 5^2 + 58^2 + 63^2 + 64^2 + 65^2 + 66^2

2nd: 4^2 + 6^2 + 7^2 + 47^2 + 48^2 + 53^2 + 56^2 + 57^2 + 79^2

3rd: 9^2 + 10^2 + 11^2 + 12^2 + 54^2 + 55^2 + 61^2 + 70^2 + 71^2

4th: 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 60^2 + 68^2 + 69^2 + 77^2

5th: 8^2 + 18^2 + 19^2 + 21^2 + 22^2 + 59^2 + 62^2 + 67^2 + 81^2

6th: 23^2 + 24^2 + 25^2 + 26^2 + 45^2 + 49^2 + 52^2 + 72^2 + 73^2

7th: 37^2 + 38^2 + 40^2 + 41^2 + 42^2 + 43^2 + 46^2 + 51^2 + 75^2

8th: 27^2 + 29^2 + 31^2 + 33^2 + 34^2 + 39^2 + 50^2 + 74^2 + 76^2

9th: 20^2 + 28^2 + 30^2 + 32^2 + 35^2 + 36^2 + 44^2 + 78^2 + 80^2

I suspect there are more than one possible sets and hence Algebraic solution may be difficult.

Anyways will try to put more thought into this.

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@gianlino:

I meant to say that it will be difficult to find one method which gives all the possible sets. I wrote a program which finds a set and then eliminates that set from array and then iterates this process.

I started developing algebraic method similar to what Sharmistha has done. But it does not really work well. I also checked Pauley Morph's approach but it does not work for specific range with all the different numbers.

Will think more on this but first I will have to think for what all numbers this is possible. For instance can we divide squares between 1 to n^2 into n such groups? Lets take squares between 1 to 4^2 or 16 and try to divide them in n=4 such sets. The answer seems to be NO. Why? Will need to think on this. There are two sets 1^2 + 6^2 + 9^2 + 16^2 = 374 and 3^2 + 10^2 + 11^2 + 12^2 = 374 which satisfies all the condition but then one will not be able to find any pair.

In fact if the overlapping is allowed then we have following sets giving 374 as the total with 4 elements per set but as I said this condition is hard to get fulfilled.

1, 2, 12, 15

1, 6, 9, 16

2, 8, 9, 15

3, 5, 12, 14

3, 10, 11, 12

5, 6, 12, 13

6, 7, 8, 15

7, 9, 10, 12

Will try to think more on this.

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Update:

So far nothing comes to my mind which will work. This is really tough to work out!!

This might (or might not) help. (Let Σ be the sum over i = 1 to 9)

If Σx[i] = 9s, then

(1) Σ(2s - x[i]) =Σx[i] and

(2) Σ(x[i])² = Σ(2s - x[i])²

EDIT

In other words, if you find a set {a[i]} of 9 integers whose sum is 9s and whose sum of squares is 20049, then the sum of squares of the nine integers {2s - a[i]} is also 20049. I don't know, however, if such a sum exist.

EDIT. OK now I know.

v = (-12, -11, 4, 7, 17, 50, 63, 80, 81)

has a sum of 9*31 and a ssq of 20049

So s = 31.

w = 2s - v = (74, 73, 58, 55, 45,12, -1, -18, -19)

has a sum of 9*31 and a ssq of 20049

In this case, it doesn't help because the two sums of squares have 12^2 in common. But, it might still be useful in other cases.

EDIT. Yes. This phenomenon occurs a lot in magic squares. E.G.

8 3 4

1 5 9

6 7 2

Note that 8²+3²+4² = 6²+7²+2²

and 8²+1²+6² = 4²+9²+2²

There is no guarantee however that this process will produce completely different numbers nor that it will produce numbers between 1 and 81.

More generally, you might want to think about what transformations of an integer vector produce another integer vector of the same length.

Checking this idea out with V's answer, I see that it isn't very useful.

This is more adaptable to a trial and error method type solution I feel since these could be random numbers whose sum of squares are sought. Tried to break up in three squares addition summing to 41*163 and then paired up three such pairs to add upto 3*41*163.

Got the following series

1) 7^2+10^2+15^2+32^2+43^2+68^2+72^2+75^2+37^2

2) 2^2+9^2+14^2+17^2+35^2+42^2+69^2+73^2+80^2

3) .....................................................................................

May try out more in this manner as it is becoming tedious due to lack of a definite pattern. But feel this is correct directionally. Please let me know of any easier method. Thx

...apologies...I just typed in a crazy proof and found 5 minutes later that it does not work because a numerical computation error...I would delete my answer in total, but I leave open the remote possibility that I will have something intelligent to say later tonite.

17 and a half.

17 and a half

math