Let $Q(N;q,a)$ be the number of squares in the arithmetic progression $qn+a$, for $n=0$,$1,\ldots,N-1$, and let $Q(N)$ be the maximum of $Q(N;q,a)$ over all non-trivial arithmetic progressions $qn + a$. Rudin’s conjecture claims that $Q(N)=O(\sqrt{N})$, and in its stronger form that $Q(N)=Q(N;24,1)$ if $N\ge 6$. We prove the conjecture above for $6\le N\le 52$. We even prove that the arithmetic progression $24n+1$ is the only one, up to equivalence, that contains $Q(N)$ squares for the values of $N$ such that $Q(N)$ increases, for $7\le N\le 52$ ($N=8,13,16,23,27,36,41$ and $52$).
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