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Prove that the product of 4 consecutive integers is always equal to 1 less than a perfect square

Note by Vladimir Smith 2 years, 2 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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\(Let\quad x\quad be\quad equal\quad to\quad the\quad lowest\quad of\quad the\quad consecutive\quad integers.\quad Then:\\ x(x+1)(x+2)(x+3)\quad =\quad y^{ 2 }-1\\ Let\quad a\quad =\quad x(x+3)\\ Let\quad b\quad =\quad (x+1)(x+2)\\ Then\quad ab=y^{ 2 }-1\\ Lets\quad find\quad the\quad difference\quad between\quad a\quad and\quad b.\\ a+c=b\\ { x }^{ 2 }+3x+c\quad ={ \quad x }^{ 2 }+3x+2\\ \quad \quad \quad \quad \quad \quad c\quad =\quad 2\\ \therefore \\ \quad \quad a(a+2)\quad =\quad y^{ 2 }-1\\ { a }^{ 2 }+2a+1\quad =\quad { y }^{ 2 }\\ \quad \quad { (a+1) }^{ 2 }\quad =\quad { y }^{ 2 }\)Log in to reply