Mirror Squared

Firstly, let $d$ be the single digit $1 \leq d \leq 9$ that the square number is composed of. $d$ can then only be 1, 4, 5, 6, or 9 because of the last digit of square numbers.

Secondly, assume $b^{2}$ is a square number with the wanted properties: $\begin{aligned} b^{2} &= d \times 10^{n} + d \times 10^{n-1} + ... + d \times 10 + d \quad , \quad n \geq 0\\ b^{2} &= d(10^{n} + 10^{n-1} + ... + 10 + 1) \\ b^{2} &= \frac{d(10^{n+1}-1)}{9} \end{aligned}$

Case 1: $n > 0$ .
Considering quadratic residues modulo 4, we have $b^{2} \equiv 0, 1 \pmod 4$ . Since $\frac{(10^{n+1}-1)}{9} \equiv 3 \pmod 4$ , this means that we must have $d \equiv 0, 3 \pmod 4$ . Hence, $d = 3, 4, 7, \text{or} \ 8$ . Referencing the possible last digits of a square, this means that we can only have $d = 4$ . Notice that since $\frac{(10^{n+1}-1)}{9} \equiv 3 \pmod 4$ , it is not a square number. Hence $4 \times \frac{(10^{n+1}-1)}{9} \equiv 3 \pmod 4$ is also not a perfect square. Thus, there are no solutions in this case.

Case 2: $n=0$
If $n = 0$ , then we are looking at single digit numbers. The only possibilities for $d$ are 0, 1, 4, and 9; the square numbers of one digit.

So there are 4 square numbers with all of their digits equal.

The ans is:0,1,4,9 which is 0^2,1^2,2^2,3^2

this question is ambiguous.. "How many perfect squares have the property that all of their digits are equal?"....digits are "equal"????shouldnt it be digits are the "same"???

0, 1, 4 and 9.

When all digits are same, the number is
AAAA.......n times, and it is also a perfect square. Which means A×1111...n times is a perfect square. But 111....n times is equal to, = 1+10 +10^2+.......+10^(n-1)=(10^n -1)/9.
This implies that A as well as this sum should each be a perfect square separately.
Let (10^n - 1)/9 = b^2,
then 10^n - 9b^2 = 1,
=> (root 10^n + 3b )(root 10^n - 3b)= 1×1
=> n = 0 and b = 0, which mean that the number is A×(10^0)= A, a single digit
number.
So, A = 0, 1, 4 and 9 are possible answers.

I need some help. I guessed and got it right, but I definitely do not understand this question at all. Which to me, makes my answer wrong. What exactly is a perfect square?? And what is meant by "have the property that all of their digits are equal"?? Thanks for the help.

A number of the same digits, as seen in other solutions, must be divisible by a number made of only 1s. However, these numbers can only be divisible by themselves, 1, and the set of 3 and the 037 string if the number of 1s is divisible by 3.

However, the 1 string has to be multiplied by itself in order to have this be a perfect square. Thus, the only 1 string that fits is 1 itself - so they have to be 1 digit perfect squares.

The positive single-digit perfect squares are 1, 4, and 9. However, 0 is also a perfect square too: $0^{2}=0$

How about 7 as an answer? Is (-3)^2 a different square than (+3)^2 They both result in +9, but are they different squares?

Then $a^2<10$ or $a<\sqrt{10}$ that $a$ is a integer which is $0,1,2,3$

The solution is to prove that no same digit serie with a length > 1 can be a perfect square. If it was it'll be obviously something like 111111... , 444444.... or 9999..... because of 1, 2^2, 3^3 multiplying 11111... So we have to prove that any serie of 1 cannot be a perfect square. It s obvious for 11 as a prime number so what about length > 3. Lets analyse the 2 last digits of the root of this hypothetical square. It would be something like A1 or A9 because only 1 and 9 squares ends by 1. the squares of such roots should have thoses patterns on their 2 last digits : - A1 : 2 A ;1 : 2 A cannot end by 1 (even) - A9 :18 A +8;1 : 18 A +8 cannot end by 1 - If it was the case 18*A should end by 3 - Impossible So the solution is 0,1,4 and 9

Since, there is no such no. greater than equal to 2 digits having ALL the digits same and the number being a perfect square, we need to choose only 1 digit perfect squares. Those are 0= $0^{2}$ , 1= $1^{2}$ , 4= $2^{2}$ , 9= $3^{2}$ . Thus, these 4 digits are the required answers.

It could be thought as since the digits are to be same then all the single digit perfect square(0,1,4,9) would be the answer plus we have to check whether any unit multiple (i.e 0-9) of numbers like 11,111,1111 and similarly 1111111...... would be perfect square and it is simple to see they won't be.

What if i say -1,-2,-3 are also the solutions

All square integers end in either 0, 1,4,9,6,5. So if there is a square S with all digits equal it must be dddd..d where d is one of 0,1,4,5,6,9. Further if the number has n decimal places and d is 1,4, or 9 easy to see that S/d is a square with all its digits equal to 1. On the other hand if d=6, 6^2 must divide S and so 6 must divide S/d which is odd, a contradiction. If d=5, 5^2 divides S and so 5 divides S/d, that is impossible because S/d ends in 1 and any number divisible by 5 must end in either 5 or 0. The question is reduced to finding all squares whose digits are all equal to 1. In this case if S has n digits, S=1+ 10^1+.......+10^(n-1) which is equal to (10^n-1)/9= (say) b^2. Thus (10^n-1)=(3b)^2 which is congruent to 1 mod 4 because 3b is odd. Thus 10^n is congruent to 2 mod 4. That is impossible unless n=1. This means S=1. Hence the only possibilities are 0,1,4,9.

$\overline{xxx...x} = 111...1x$ , $0 \le x \le 9$ .

When number is higher than 1 digit, we need 111...1 to be equal to x for the number to be a perfect square. However, since x can not be higher than 9, we are looking for one digit numbers.

Next we check for the perfect squares lesser or equal to 9. There are $\boxed{4}$ such ones.