Determine all possible 7-digit decimal (base 10) perfect square(s), each of whose digits is nonzero and even.

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

The root of a 7-digit perfect square is 4 digits (inclusively between 1000 to 3162). Representing the digits of the root as ABCD, ABCD² is equal to

10⁶*(A²) +
10⁵*(2AB) +
10⁴*(2AC + B²) +
10³*(2AD + 2BC) +
10²*(2BD + C²) +
10¹*(2CD) +
10⁰*(D²)


As the final digit must be non-zero and even, D must be 2 or 8; and the 10⁰-digit digit must be 4. Proceeding to the next, if D was 2 then C must be 2 or 6 with respective values for the 10¹-digit as 8 or 4; and if D was 8 then C must be 0, 3, 6 or 7 with respective values for the 10¹-digit as 6, 4, 2 or 8. By substituting each of the ten possible digits for B and one of the three (1-3) for A, eliminating any number where any one of the digits is zero or odd, one can find each digit of the 7-digit perfect square and it's root, ABCD. Only three solutions are found:
2862864 (1692²), 8282884 (2878²), and 8868484 (2978²).

Using the above procedure manually may be onerous, but it may be done using simple math and without the aid of a computer program.

Posted by Dej Mar on 2008-10-27 04:12:05