Prove that for any natural number \(x\), \( x(x+1)(x+2)(x+3)+1 \) is a perfect square.

Prove that for any natural number \(x\), \( x(x+1)(x+2)(x+3)+1 \) is a perfect square.

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TopNewestI don't think this a legit proof but here it is.

the expression can be written as

\(\large (x^{2}+3x)(x^{2}+3x+2)\) \( \text{Now Substitute} \hspace{1mm}(\large x^{2}+3x) =y\).

The above expression changes to \(\large (y)(y+2)\) = \(\large y^{2}+2y\)

On Adding 1 gives us \(\large y^{2}+2y+1\)= \((y+1)^{2}\)

\(\large QED\) – Achal Jain · 5 months, 1 week ago

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Note that you have to show that y+1 is an integer, so that the number is a perfect square (as opposed to a perfect square as an expression). – Calvin Lin Staff · 5 months, 1 week ago

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it is mentioned that \(x\) is a natural number \(\therefore\) \(x^{2}+3x\) must also be an integer. This proves that \( y\) must also be an integer.

Does this complete the proof? – Achal Jain · 5 months, 1 week ago

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(likely just a typo) Note that you want \( x^2 + 3x = y \). – Calvin Lin Staff · 5 months, 1 week ago

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