Product of 4 consecutive integers

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

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Solution posted here

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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 }

Vladimir Smith - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$