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U3A MATHEMATICAL PUZZLES
Proof that odd numbers consisting only of 1s are never perfect squares.
A perfect square is the result of multiplying a whole number by itself. A number can
be dissembled into two parts, A and B, such that A represent the whole tens, and B the
digits less than ten. E.g
54321 can be written 54320 + 1, where A is 54320 and B is 1.
The square can thus be written as 543212 or as (54320 + 1)2 or (A + B)2
(A + B)2 can be expanded to A2 + 2AB + B2 or 2950662400 + 108640 + 1
Note that the last digit of the sum must always be the last digit of B2
Since the only digits which square to 1 are 1 and 9, B must be 1 or 9.
If B is 1, then A2 + 2AB = A2 + 2A
If B is 9, then A2 + 2AB = A2 + 18A
= A(A + 2)
= A(A + 18)
Since A is a multiple of 10 the product of A(A+2) must also be a multiple of 10 and
the sum of (A+2) must also be even: that is the tens digit must always be even. In this
case B2 is 1, so the addition of B2 to A(A+2) only affects the last digit. Thus it is not
possible to have a perfect square ending in …11 when B is 1.
Similarly, when B is 9, the product of A(A+18) is a multiple of 10 and the sum of
A+18 is always even. Thus the tens digit must always be even. B2 = 81, so when the
tens digit has 8 added to it, it remains even. Thus it is not possible to have a perfect
square ending in …11 when B is 9.
Hence a number consisting entirely of 1s can never be a perfect square.
Chris Petrie
16.11.08