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 Base N Perfect Square Poser (Posted on 2016-07-13)
Consider a base-N nonzero perfect square of the form X1X1X2X2, where X1 and X2 are base-N digits whether same or different.

Derive a general formula that expresses X1X1X2X2 in terms of N, where N is an integer > 3.

*** X1X1X2X2 is a concatenation of the digits and not the product.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution Comment 2 of 2 |
I am going to use XXYY as the number to avoid subscripts.

XXYY is a multiple of N+1, XXYY = (N+1)*(X0Y).  Because XXYY is a square it must be a multiple of (N+1)^2.  This imples X+Y=N+1

Also, (N+1)*(AB) = X0Y, which implies AB0+AB=X0Y.  Then A=X-1 and B=Y.  If B=4 and A=N-4 then AB=N*(N-4)+4 = (N-2)^2.  This is a perfect square.  Therefore for any base N>=5, (X,Y)=(N-3,4) will produce a desired perfect square whose value will equal [(N-2)*(N+1)]^2.

But these are not the only squares.  Substituting expressions for X and Y into perfect square AB yields (X-1)^2+(X-1)*Y+Y.  The smallest nontrivial perfect square occurs when X=3 and Y=7, implying 3377 base 9 is a perfect square, 3377 base 9 = 2500 = 50^2.

Further nontrivial solutions include (using A=10, B=11, etc.) 44AA base 11, 8899 base 16, 33FF base 17, 55DD base 17, 44II base 21, 55GG base 21, etc.

 Posted by Brian Smith on 2016-07-14 09:44:30

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