<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] I tried solving came out with following results:
$(i)-(ii)$
gives
$a+b+c=0$
which implies,
$a^3+b^3+c^3=3abc$
Multiplying equation $(i)$ with $a$ and multiplying equation $(ii)$ with $b$
and then subtracting $(i)-(ii)$
we get 
$a^2+b^2+ab=1$
which can be written as 
$(a+b)^2-ab=1$ , which can be further written as $c^2-ab=1$. 
</s>