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We use these results to show that: (a) A latin square of order $n$ cannot have more than $\\frac nm{n \\choose h}/{m\\choose h}$ subsquares of order $m$, where $h=\\lceil(m+1)/2\\rceil$. Indeed, the number of subsquares of order $m$ is bounded by a polynomial of degree at most $\\sqrt{2m}+2$ in $n$. 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Wanless, Joshua M. Browning, Petr Vojt\\v{e}chovsk\\'y","submitted_at":"2015-09-18T15:30:01Z","abstract_excerpt":"We derive necessary and sufficient conditions for there to exist a latin square of order $n$ containing two subsquares of order $a$ and $b$ that intersect in a subsquare of order $c$. We also solve the case of two disjoint subsquares. We use these results to show that: (a) A latin square of order $n$ cannot have more than $\\frac nm{n \\choose h}/{m\\choose h}$ subsquares of order $m$, where $h=\\lceil(m+1)/2\\rceil$. Indeed, the number of subsquares of order $m$ is bounded by a polynomial of degree at most $\\sqrt{2m}+2$ in $n$. 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