A perfect square

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

Note by Akarsh Jain
1 year, 3 months ago

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I 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 - 1 year, 3 months ago

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That works!

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 - 1 year, 3 months ago

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Thanks

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 - 1 year, 3 months ago

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@Achal Jain Yes.

(likely just a typo) Note that you want \( x^2 + 3x = y \).

Calvin Lin Staff - 1 year, 3 months ago

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