Q: Prove that $a = b$ (Leningrad 1990) Let a and b be natural numbers that $b ^ 2 + ba + 1$ divides $a ^ 2 + ab +1$. Prove that $a = b$
I thought of the limitation property $b ^ 2 + ba + 1 ≤ a ^ 2 + ab + 1$ and hence $b ≤ a$, but I don't know how to process. How can I prove it differently?

A: If $b^2 + ba + 1$ divides $a^2 + ab+1$, then it also divides $(a^2+ab+1)-(b^2+ba+1) = a^2-b^2$, which factors as $(a+b)(a-b)$.
However, it can't share any divisors with the first factor: $\gcd(b^2+ba+1, a+b) = \gcd(b^2+ba+1 - b(a+b), a+b) = \gcd(1, a+b) = 1$.
Therefore $b^2+ba+1$ also divides $\frac{a^2-b^2}{a+b} = a-b$.
However, $b^2+ba+1 > a > a-b$ (assuming $a,b\ge 1$) so the only way this can hold is if $a-b=0$, or $a=b$.
(If we allow $b=0$, then $a$ can be anything and $b^2+ba+1 =1$ still divides $a^2+ab+1=a^2+1$, so it appears our definition of "natural number" excludes $0$ here.)

A: If $b^2+ba+1$ divides $a^2+ab+1$ then $b^2+ba+1 \le a^2+ab+1$ and so $b \le a$.
Suppose $b<a$. Then $b^2+ba+1$ divides the difference $(a^2+ab+1) - (b^2+ba+1) = a^2-b^2$.
But $b^2+ba+1 = b(a+b)+1 = 1 \mod a+b$ and $a^2-b^2=(a-b)(a+b) = 0 \mod a+b$.
So either $a+b=1$, in which case $b=0, a=1$; or $b \not \lt a$ in which case $a=b$.
