Question Number 212584 by MrGaster last updated on 18/Oct/24
Commented by mr W last updated on 18/Oct/24
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Commented by MrGaster last updated on 18/Oct/24
I can't modify and divide this post by three, so I will state the topic in the form of comments:
At least how many squares (including 0) sum can represent all natural numbers. If you choose at random, it seems that 4 is the final answer, and most of them are 8n+7. Now please prove that 8n+7 cannot be written as the sum of three squares
2.Excluding these 8n+7, there are still four numbers left, and I find that they are related to 8n+7: they can be written in the form of 4 a * (8n+7). Please prove that all 4 a* (8n+7) cannot be written in the form of the sum of at least three squares.
Answered by Frix last updated on 18/Oct/24
$$\mathrm{8}{n}+\mathrm{7}={u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}^{\mathrm{2}} \\ $$$$\mathrm{8}{n}+\mathrm{7}\:\mathrm{is}\:\mathrm{odd}\:\Rightarrow\:\begin{cases}{{u}=\mathrm{2}{j}+\mathrm{1}\wedge{v}=\mathrm{2}{k}\wedge{w}=\mathrm{2}{l}}\\{{u}=\mathrm{2}{j}+\mathrm{1}\wedge{v}=\mathrm{2}{k}+\mathrm{1}\wedge{w}=\mathrm{2}{k}+\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\mathrm{8}{n}+\mathrm{7}=\left(\mathrm{2}{j}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{k}\right)^{\mathrm{2}} +\left(\mathrm{2}{l}\right)^{\mathrm{2}} \\ $$$$\mathrm{8}{n}+\mathrm{6}=\mathrm{4}{j}^{\mathrm{2}} +\mathrm{4}{j}+\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{l}^{\mathrm{2}} \\ $$$$\underset{\mathrm{odd}} {\underbrace{\mathrm{4}{n}+\mathrm{3}}}=\underset{\mathrm{even}} {\underbrace{\mathrm{2}\left({j}^{\mathrm{2}} +{j}+{k}^{\mathrm{2}} +{l}^{\mathrm{2}} \right)}} \\ $$$$\mathrm{impossible}\: \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{8}{n}+\mathrm{7}=\left(\mathrm{2}{j}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{l}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{8}{n}+\mathrm{4}=\mathrm{4}\left({j}^{\mathrm{2}} +{j}+{k}^{\mathrm{2}} +{k}+{l}^{\mathrm{2}} +{l}\right) \\ $$$$\underset{\mathrm{odd}} {\underbrace{\mathrm{2}{n}+\mathrm{1}}}=\underset{\mathrm{even}} {\underbrace{{j}\left({j}+\mathrm{1}\right)}}+\underset{\mathrm{even}} {\underbrace{{k}\left({k}+\mathrm{1}\right)}}+\underset{\mathrm{even}} {\underbrace{{l}\left({l}+\mathrm{1}\right)}} \\ $$$$\mathrm{impossible} \\ $$
Commented by MrGaster last updated on 18/Oct/24
$$ \\ $$Excluding these 8n+7, there are still four numbers left, and I find that they are related to 8n+7: they can be written in the form of 4 a * (8n+7). Please prove that all 4 a* (8n+7) cannot be written in the form of the sum of at least three squares.