Prove that $\frac{2022}{n}+4n$ is a perfect square iff $\frac{2022}{n}-8n$ is a perfect square

Charlie Conner
2022-10-06
Answered

Prove that $\frac{2022}{n}+4n$ is a perfect square iff $\frac{2022}{n}-8n$ is a perfect square

You can still ask an expert for help

omeopata25

Answered 2022-10-07
Author has **5** answers

You can greatly reduce the number of cases to check by working modulo $4$. We have

$2022=2\cdot 3\cdot 337\equiv 2\cdot 3\cdot 1\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}4).$

Since a square must be $0$ or $1$ mod $4$, we can immediately reduce to the two cases $n\in \{6,2022\}$. And of course $n=2022$ is impossible since $\frac{2022}{n}-8n$ will be negative.

$2022=2\cdot 3\cdot 337\equiv 2\cdot 3\cdot 1\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}4).$

Since a square must be $0$ or $1$ mod $4$, we can immediately reduce to the two cases $n\in \{6,2022\}$. And of course $n=2022$ is impossible since $\frac{2022}{n}-8n$ will be negative.

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We have

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Do we have a proof relying on rudimentary techniques?

We have

$\sum _{j=1}^{n}\frac{1}{{Q}^{\prime}({\alpha}_{j})}=0.$

Do we have a proof relying on rudimentary techniques?

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Premise $1$: $P\to \mathrm{\neg}R$

Premise $2$: $Q\to S$

Premise $3$: $(R\vee S)\to T$

Premise $4$: $\mathrm{\neg}T$

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Premise $1$: $P\to \mathrm{\neg}R$

Premise $2$: $Q\to S$

Premise $3$: $(R\vee S)\to T$

Premise $4$: $\mathrm{\neg}T$

Hypothesis: $P\vee Q$

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