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

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

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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 · 1 year, 11 months agoLog in to reply