<s>[INST] find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the condition $a+b=-c$ are such that $a^2=bc+1$ and $b^2=ca+1$. Also if $a-b=b-a,$ it means $a=b.$ Then both $b-a$ and $a-b$ are $0.$ Hence $c$ can thus be any integer. Hence, there are two groups of integers satisfying the required condition $-a=b$ with any $c$ and $a+b=-c.$
Please let me know whether my solution is correct or not.
 [/INST] Assuming $b\neq a$ you gained a constraint
$$-(a+b) = c$$
This constriction is not bad, because there is no number $b = c$ such that $a^2 = bc + 1$ for integers.
It's fine to me.
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