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I have a multiset A of $[n]$, $|A| = n$ and $A\not = [n]$ (i.e., A has at least one element repited) whose sum is equal $\frac{n(n+1)}{2}$, i.e.,

$$\sum_{a\in A} a = \frac{n(n+1)}{2}$$

Can it's square sum $\sum_{a\in A}a^2$ be equal to $\frac{n(n+1)(2n+1)}{6}$?

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1 Answers1

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Searching for numbers that can be written in many ways as a sum of a certain number of squares produces results quickly. I imagine there are very many qualifying results.

Using the result for $129$ noted in Determining the number of ways a number can be written as sum of three squares, $$10^2+5^2+2^2=8^2+8^2+1^2=129$$

and noting $10+5+2=8+8+1$, it's easy to see that your property holds for the multiset $\{1,1,3,4,6,7,8,8,8,9,11\}$ (by substituting $\{1,8,8\}$ for $\{2,5,10\}$ in $[11]$).

I "went to eleven" here just so that $n$ was included in $A$.

A fairly simple program search for numbers that can be represented as a sum of three positive squares multiple ways with common sums finds the smallest solution that has at least one of the solutions being composed of distinct numbers as $62 = 1^2+ 5^2 +6^2 = 2^2 + 3^2 + 7^2$ so the least $n$ is probably $7$.

results to $200$:

sq total 62 sum 12 (1, 5, 6) (2, 3, 7)
sq total 89 sum 15 (2, 6, 7) (3, 4, 8)
sq total 101 sum 15 (1, 6, 8) (2, 4, 9)
sq total 118 sum 16 (1, 6, 9) (3, 3, 10)
sq total 122 sum 18 (3, 7, 8) (4, 5, 9)
sq total 129 sum 17 (1, 8, 8) (2, 5, 10)
sq total 134 sum 18 (2, 7, 9) (3, 5, 10)
sq total 146 sum 18 (1, 8, 9) (3, 4, 11)
sq total 150 sum 18 (1, 7, 10) (2, 5, 11)
sq total 153 sum 19 (2, 7, 10) (4, 4, 11)
sq total 161 sum 21 (4, 8, 9) (5, 6, 10)
sq total 166 sum 20 (2, 9, 9) (3, 6, 11)
sq total 173 sum 21 (3, 8, 10) (4, 6, 11)
sq total 185 sum 21 (2, 9, 10) (4, 5, 12)
sq total 189 sum 21 (2, 8, 11) (3, 6, 12)
sq total 194 sum 20 (1, 7, 12) (3, 4, 13)
sq total 194 sum 22 (3, 8, 11) (5, 5, 12)

OEIS A282241 is (I think) a subset of these values.

This property becomes common for larger numbers; $6515$ of the numbers up to $10000$ can be expressed as a sum of three squares in two or more ways where the square roots also have a common sum.

Python code (improved to get to higher numbers):

lim = int(input('Search limit: '))
rlim = int(lim**0.5)
sqval = {i:i*i for i in range(rlim+1)}

sieve


allz = []
for a in range(1,rlim+1):
    for b in range(a,rlim+1):
        ab = sqval[a]+sqval[b]
        if ab > lim:
            break
        for c in range(b, rlim+1):
            abc = ab+sqval[c]
            if abc > lim:
                break
            allz.append( (abc, a+b+c, a,b,c) )


print(len(allz))


allz.sort()
allz.append( (0,0,0,0,0) ) # force action on last totals


pM = 0
pt = 0
mz = []
qct = 0
Mql = 0
for M,t,a,b,c in allz:
    if M != pM or t != pt:
        # report previous block if sutiable
        if len(mz) > 1:
            if any(len(set(mzi))==3 for mzi in mz):
                # report suitable results
                Mql = 1
                print("sq total", pM, "sum", pt, *mz)
        mz = []
        pt = t
        if M != pM:
            qct += Mql
            pM = M
            Mql = 0
    mz.append( (a,b,c) )


print (qct, "numbers in range have a suitable representation")
Joffan
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