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===Gardner square=== The Gardner square, named after recreational mathematician [[Martin Gardner]], similar to the Parker square, is given as a problem to determine a, b, c and d.{{Citation needed|date=January 2024}} {| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;table-layout:fixed;width:8em;height:8em;" |- | 127{{sup|2}}|| 46{{sup|2}}|| 58{{sup|2}} |- | 2{{sup|2}}|| b{{sup|2}}|| c{{sup|2}} |- | a{{sup|2}}|| 82{{sup|2}}|| d{{sup|2}} |} This solution for a = 74, b = 113, c = 94 and d = 97 gives a semimagic square; the diagonal {{math|127{{sup|2}} + b{{sup|2}} + d{{sup|2}}}} sums to {{val|38307}}, not {{val|21609}} as for all the other rows and columns, and the other diagonal.<ref>{{cite journal |last1=Gardner |first1=Martin |title=The magic of 3x3 |journal=Quantum |date=January 1996 |volume=6 |issue=3 |pages=24β26 |url=https://static.nsta.org/pdfs/QuantumV6N3.pdf |access-date=6 January 2024 |issn=1048-8820}}</ref><ref>{{cite journal |last1=Gardner |first1=Martin |title=The latest magic |journal=Quantum |date=March 1996 |volume=6 |issue=4 |page=60 |url=https://static.nsta.org/pdfs/QuantumV6N4.pdf |access-date=6 January 2024 |issn=1048-8820}}</ref><ref>{{cite journal |last1=Boyer |first1=Christian |title=Some Notes on the Magic Squares of Squares Problem |journal=The Mathematical Intelligencer |date=12 November 2008 |volume=27 |issue=2 |pages=52β64 |doi=10.1007/BF02985794}}</ref> {| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;table-layout:fixed;width:19em;height:8em;border-style:none;" |- | style="background-color: white; border-style: none;" | || 127{{sup|2}}|| 46{{sup|2}}|| 58{{sup|2}} || 21609 |- | style="background-color: white; border-style: none;" | || 2{{sup|2}} || 113{{sup|2}}|| 94{{sup|2}} || 21609 |- | style="background-color: white; border-style: none;" | || 74{{sup|2}} || 82{{sup|2}}|| 97{{sup|2}} || 21609 |- | 21609 || 21609 || 21609 || 21609 || 38307 |}
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