Find all integer $a,b,c$ such that $$a^2=bc+4 \qquad b^2=ac+4$$
My work $$a^2-b^2=c(b-a)$$ $$(a-b)(a+b)=-c(a-b)$$ $$(a+b)=-c \qquad \text{if}\space a\ne b $$
I don't know how to proceed ahead .
Find all integer $a,b,c$ such that $$a^2=bc+4 \qquad b^2=ac+4$$
My work $$a^2-b^2=c(b-a)$$ $$(a-b)(a+b)=-c(a-b)$$ $$(a+b)=-c \qquad \text{if}\space a\ne b $$
I don't know how to proceed ahead .
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Case 1: $a=b$. So $a^2=ac+4$, which means $a(a-c)=4$. Then $a=\pm 1,\pm 2,\pm 4$.
Case 2: $a\neq b$. As you wrote, $c=-(b+a)$, which means $a^2+ab+b^2=4$. Multiplying by $4$, we get $(2a+b)^2+3b^2=16$.
Can you complete it from here?