Question

find the number of triplets of all integers (a,b,c) such that a^2=bc+1, b^2=ac+1​

Answer 1

Answer:

$a2 = bc + 1 \\ b2 = ac + 1$

Answer 2

Step-by-step explanation:

$\huge\mathcal{Given}\\a^2=bc+1\rightarrow{EQ:01}\\b^2=ac+1\rightarrow{EQ:02}\\a^2-bc=1\\b^2-ac=1\\a^2-bc=b^2-ac\\(a+b)(a-b)=c(b-a)\\(a-b)(a+b+c)=0\\Case\:1\:a-b=0\\a=b\\From\:eq:01\\a^2-ac=1\\a(a-c)=1\\If\:a\leq\:c,\:yields\:either\:negative\:or\:zero\\So,\:a\:must\:be\:greater\:than\:c\\Another\:observation\:that\:if\:a>2\:leads\:higher\:result\:than\:1\\It's\:quite\:interesting\:to\:eliminate\:like\:so\:only\:in\:case\:of\:integer\:solutions.\\So,clearly\:c=0\:holds\:that\:equations\:results\\a^2=1\rightarrow\:a=\pm1;\:b=\pm1\\The\:solution\:set\:is\{\pm1,\pm1,0\}\\Case\:2\:a+b+c=0=>c=-(a+b)\\From\:Eq\:01\\a^2=b(-(a+b))+1\\a^2=-ab-b^2+1\\a^2+b^2+ab-1=0\\a^2+ab+(b^2-1)=0\\A\:quadratic\:equation\:in\:terms\:of\:a\\a=\frac{-b\pm\sqrt{b^2-4.1.(b^2-1)}}{2}\\a=\frac{-b\pm\sqrt{4-3b^2}}{2}\\To\:get\:an\:integer\:value\:for\:a\\\sqrt{4-3b^2}\:must\:be\:an\:integer\\implies\:4-3b^2\geq\:0\:and\:must\:be\:an\:integer\\For\:b\:\epsilon\:[-1,1]\:results\:real\:values\\At\:values\:of\:b=-1,0,1\:a\:yields\:6\:values\:results\:six\:solutions\\So,\:in\:total\:eight\:integer\:solutions\:are\:possible\\(a,b,c)\:\epsilon\:\\\{(1,1,0)\\\:(-1,-1,0)\\\:(1,-1,0)\\\:(0,-1,1)\\\:(1,0,-1)\\\:(-1,0,1)\:\\(0,1,-1)\\\:(-1,1,0)\}$

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