# Multiple odd squares sum to an odd square

Any number of appropriately chosen squares can sum to a square. However, this is not the case when the square addends and the square sum are all required to be odd. Since all odd squares are $$\equiv 1 \bmod 8$$, only sums of $$8k+1$$ odd squares can equal an odd square. To be clear, it should be possible to find collections of $$9,17,\text{ or }25$$ (etc.) squares that sum to an odd square, but not collections of (for example) $$4$$ or $$15$$ squares that do so. For $$a_i,b$$ odd $$\sum_{i=1}^m a_i^2=b^2$$ can have solutions only if $$m=8k+1$$

If $$k$$ is chosen to be a triangular number $$k=\frac{n(n+1)}{2}$$ then $$8k+1=(2n+1)^2$$ and trivial solutions of the form $$a_i=c$$ arise where the sum is $$b^2=((2n+1)c)^2$$

In the case of $$k=9$$ it is not hard to identify collections of squares that satisfy the conditions if the same square summand can be used more than once. For example, $$1+1+9+9+9+9+9+25+49=121$$ and $$1+1+1+1+9+9+25+25+49=121$$. More interesting would be cases in which the nine square summands were different. The closest such instances I have found by hand computation are $$9+25+49+81+121+121+169+225+289=1089=33^2$$ and $$9+25+49+81+121+121+169+289+361=1225=35^2$$. Each of these has only one duplicated square, $$121$$. Since the first nine odd squares sum to $$969$$, it will be the case that any collections featuring nine distinct squares will require $$b>33$$

Although I have no reason to doubt that examples with no duplications can be found (and I would deem it of interest to see some), such as by computer searches, my real question is: Are there algorithmic methods by which such collections of odd squares can be generated or identified?

• So , you are looking for distinct odd squares summing up to another odd square ? Aug 13, 2023 at 17:36
• $$1^2+3^2+5^2+7^2+9^2+11^2+13^2+15^2+29^2=39^2$$ Aug 13, 2023 at 17:45

Let $$x_1, x_2, \ldots x_{8k}$$ be an odd numbers, $$N = \sum_{i = 1}^{8k}x_i^2$$. $$\frac{N + 4}{4}$$ and $$\frac{N - 4}{4}$$ is an odd numbers since $$N$$ is divisible by 8. $$N = \left(\frac{N + 4}{4}\right) ^2 - \left(\frac{N - 4}{4}\right)^2$$ $$\left(\frac{N + 4}{4}\right) ^2 = \sum_{i = 1}^{8k}x_i^2 + \left(\frac{N - 4}{4}\right)^2$$ If $$x_i$$ are pairwise distinct, then we only need to check that $$x_i \ne \frac{N - 4}{4}$$. This is true because $$\left(\frac{N + 4}{4}\right) ^2 < 2\left(\frac{N - 4}{4}\right)^2$$ for $$N > 25$$.
• I think this approach can be used for any $8k+1$ collection of squares, viz: $N=\sum_{i=1}^{8k}x_i^2$ and use same logic as presented in the answer. Maybe add that to your answer if you think it is correct. Aug 14, 2023 at 0:51
• I replaced 8 with $8k$. Aug 14, 2023 at 7:17