0, 25, 2025, 13225…are squares that remain squares
if every digit in the number defining them is augmented by 1.

Let's call them **squarish** numbers.

a. List two more samples of squarish numbers.

b. Prove that all such numbers are evenly divisible by 25.

c. Why are there neither 3-digit nor 6-digit squarish numbers?

d. Prove that between 10^k and 10^(k+1) there is at most one squarish number.

Well, I am not clear what it means to augment every digit by 1.

Consider 89. Does augmenting every digit by 1 turn 89 into 90 or into 100? I'll assume no carryover, and thus 89 augmented is 90. That's the way my bicycle cylinder lock works.

Having said that, all squares > 9 end in one of the following pairs of ending digits. Only one of the augmented ending digits is itself a possible ending pair of digits (36), which means that all squarish numbers greater than 9 must end in 25. The only squarish number less than 9 is 0, which is a multiple of 25, so all squarish numbers are multiples of 25,

ending augments to

00 11

01 12

04 15

09 10

16 27

21 32

24 35

25 **36**

29 30

**36** 47

41 52

44 55

49 50

56 67

61 72

64 75

69 70

76 87

81 92

84 95

89 90

96 07

Note that this proof needs some more work if something ending in 89 augments to something ending in 00, so my interpretation of the problem was a little self-serving.

It looks like OEIS sequence A061843 (cited by Charlie) gets around the whole issue by saying that a squarish number is not allowed to have any 9's in it. If this is Ady's definition also, then my proof is complete.

*Edited on ***March 13, 2011, 4:01 pm**

*Edited on ***March 13, 2011, 4:04 pm**