Number Theory problem

A four digit number has the following properties:

(i) it is a perfect square,

(ii) its first two digits are equal to each other,

(iii) its last two digits are equal to each other .

Find all such four-digit numbers.

Till now I have been able to find only one such number. I don't know if I am correct or not. Any idea ? Note by Nishant Sharma
6 years, 4 months ago

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Isn't this an RMO problem?

- 6 years, 4 months ago

Yep, RMO 1991 or 1992 maybe.

- 6 years, 1 month ago

Let the first two digits be $a$, and the last two digits be $b$. Then the number is $N = 1000a + 100a + 10b + b = 1100a + 11b= 11(100a+b)$. Note that $11$ divides $N$, and $11$ being a prime, for $N$ to be a perfect square $11$ must also divide $\frac{N}{11}$, i.e. $11$ must divide $100a + b$. But note that $100a + b= 99a + a+b = 9*11a + a+b$. Hence for $11$ to divide $100a + b$, $11$ must divide $a+b$, i.e. $a+b$ must be a multiple of $11$. Note that $1 \leq a \leq 9$ , $0 \leq b \leq 9$. Thus $1 \leq a+b \leq 18$. The only multiple of $11$ between $1$ and $18$ is $11$ itself. Thus $a+b= 11$. Now note that the quadratic residues modulo $10$ are $0$, $1$, $4$, $5$, $6$, and $9$. Thus the only possible values of $b$ are$4$, $5$, $6$, and $9$ (note that $b$ can't be $0$ or $1$ since that would imply $a= 11$ or $a= 10$ respectively ) .

For $b= 4$, we have $a= 7$, and the number is $7744 = 88^2$ (a perfect square). For $b= 5$ we have $a= 6$ and the number is $6655$ (not a perfect square). For $b= 6$ we have $a= 5$ and the number is $5566$ (not a perfect square). For $b= 7$ we have $a= 4$ and the number is $4477$ (not a perfect square).

Hence there is only one number that satisfies the given conditions and it is $7744$.

- 6 years, 4 months ago

This is good, though there isn't a need to appeal to quadratic residues (for those who don't know it).

Once you know that the cases are $a+b = 11$, there are 9 values to check for squares, and this can be done easily.

Alternatively if you are too lazy to find square roots, note that $\overline{aabb} = 121 \times (9a + 1 )$ (using that $a+b = 11$ ), so $9a + 1$ must be a perfect square. Checking $a = 1$ to 9, gives $a = 7$ as the only square.

Staff - 6 years, 4 months ago

I could understand everything except the "quadratic residues modulo 10" thing. Please explain what does that mean ?

- 6 years, 4 months ago

An integer $q$ is called a quadratic residue mod $n$ if and only if there exists an integer $x$ such that $x^2 ≡ q (mod n)$. To put it other way round, if a sequence of integers $r_i$ is defined such that $r_i$ is the remainder when $i^2$ is divided by $n$, then the quadratic residues mod $n$ are the distinct elements of the set $[ r_1, r_2, ..., r_{n-1} ]$.

- 6 years, 4 months ago

It means the last digit of a perfect square number can only be 0,1,4,5,6,9. You can prove it yourself, it is quite easy. Best wishes :)

- 6 years, 4 months ago

Oh I see.

- 6 years, 4 months ago