A perfect square

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

#Algebra

Note by Akarsh Jain
4 years, 3 months ago

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Comments

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

the expression can be written as

(x2+3x)(x2+3x+2)\large (x^{2}+3x)(x^{2}+3x+2) Now Substitute(x2+3x)=y \text{Now Substitute} \hspace{1mm}(\large x^{2}+3x) =y.

The above expression changes to (y)(y+2)\large (y)(y+2) = y2+2y\large y^{2}+2y

On Adding 1 gives us y2+2y+1\large y^{2}+2y+1= (y+1)2(y+1)^{2}

QED\large QED

Achal Jain - 4 years, 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 - 4 years, 3 months ago

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Thanks

it is mentioned that xx is a natural number \therefore x2+3xx^{2}+3x must also be an integer. This proves that y y must also be an integer.

Does this complete the proof?

Achal Jain - 4 years, 3 months ago

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@Achal Jain Yes.

(likely just a typo) Note that you want x2+3x=y x^2 + 3x = y .

Calvin Lin Staff - 4 years, 3 months ago
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