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Notice that the number $78$ is a three-digit palindrome when written in base $5$ since $(78)_{10} = (303)_5$; it is also a three-digit palindrome when written in base $7$ since $(78)_{10} = (141)_7$. Prove that there are infinitely many positive numbers $N$ that are three-digit palindromes to two different bases at the same time.

Let the two bases be $b_1$ and $b_2$. Then we must have $aba_{b_1} = cdc_{b_2}$ with the condition that $0 \leq a,b \leq b_1-1$ and $0 \leq c,d \leq b_2-1$. Expanding the given equation gives $$ab_1^2+bb_1+a = cb_2^2+db_2+c.$$ Thus $$a(b_1^2+1)+bb_1 = c(b_2^2+1)+db_2.$$ How can we find infinitely many solutions to this?

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There is a lot of freedom in finding these solutions. For example, you can simply restrict to looking for infinitely many solutions with $a=1$. Take any $b>3$ and set $b_1:=b+1$. Then $$ab_1^2+bb_1+a=2b^2+3b+2,$$ is a palindrome in base $b_1$ and in base $b$.

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