144...44

Every number $1 \underbrace{44 \ldots 44}_{\text{n digits}}$ can be written as $10^n + \underbrace{44 \ldots 44}_{\text{n digits}} = 10^n + 4 \cdot ( \underbrace{11 \ldots 11}_{\text{n digits}} ) = 10^n + 4 \cdot \frac {\left( 10^n-1 \right)} 9$ .

For $n \geq 2$ , we can factor out $2^2$ , since we only want to prove that the expression is not a perfect square.

$10^n + 4 \cdot \frac {\left( 10^n-1 \right)} 9 = 4 \cdot \left( 25 \cdot 10^{n-2} + \frac {\left( 10^n-1 \right)} 9 \right)$

Let's look at the remainder when dividing by $4$ . The remainder of a square number divided by $4$ is always $0$ or $1$ .

The remainder of a sum is the sum of the remainders, so the remainder of $25 \cdot 10^{n-2} + \frac {\left( 10^n-1 \right)} 9$ is the remainder of $25 \cdot 10^{n-2}$ , which is $0$ for $n \geq 4$ , plus the remainder of $\frac {\left( 10^n-1 \right)} 9 = \underbrace{11 \ldots 11}_{\text{n digits}}$

To test for divisibility by $4$ , we look at the two last digits, $11$ . They give a remainder of $3$ when divided by $4$ . But since the remainder of a perfect square is either $0$ or $1$ , this number with remainder $3$ cannot be a perfect square, so there are No other such numbers.

The numbers of this form (starting at 144) are all divisible by 4, but after dividing them by 4, the quotient mod 8 quickly stabilizes: it is 4, 1, 3, 7, 7, 7... as can easily be checked. Neither 3 nor 7 are quadratic residues mod 8, so the answer is No.

Consider the remainders of such numbers after dividing by 16. A quick check tells us that perfect squares are always congruent to $\color{#20A900} 0$ , $\color{#20A900} 1$ , $\color{#20A900} 4$ , or $\color{#20A900} 9$ modulo 16.

$\begin{aligned} 14 &\equiv \color{#D61F06}14 \ \ \color{#333333} \text{mod} \ 16 \\ 144 &\equiv \ \ \color{#20A900}0 \\ 1,444 &\equiv \ \ \color{#20A900}4 \\ 14,444 &\equiv \color{#D61F06}12 \\ 144,444 &\equiv \color{#D61F06}12 \\ 1,444,444 &\equiv \color{#D61F06}12 \end{aligned}$

Note that $10,000 = 16 \times 625$ , so every number from $14,444$ on is congruent to $4,444 \equiv \color{#D61F06}12$ mod 16, and thus not a perfect square.

[This train of thought was inspired by Henry U factoring out $2^2 = 4$ and then considering divisibility by 4.]

Since any number $1444\ldots 444$ is even, it must be the square of an even number, i.e. $1444\ldots 444 = (2k)^2 = 4k^2$ .

Thus any $1444\ldots 444$ is a multiple of $4$ .

In our case $144 = 12^2 = (2\cdot 6)^2 = 4\cdot 36$

and $1444 = 38^2 = (2\cdot 19)^2 = 4\cdot 361$ .

Now let's take a look at $1444\ldots 444$ numbers with more than two $4$ digits: by dividing any such number by $4$ we get $361\ldots 111$ , in fact

$\frac{{\color{#0000ff}{144}}4\ldots 444}{4} = \frac{{\color{#0000ff}{1440\ldots 000}}}{4} + \frac{4\ldots 444}{4} = {\color{#0000ff}{360\ldots 000}} + 1\ldots 111 = {\color{#0000ff}{36}} 1\ldots 111$ .

Any $361\ldots 111$ is an odd number, thus it must be the square of an odd number, i.e. $361\ldots 111 = (2l+1)^2 = 4l^2 + 4l +1$ .

By rearranging we see that $361\ldots 110 = 4l^2 + 4l = 4(l^2 + l)$

i.e. any number on the left must be a multiple of $4$ ; but any such number must be divisible by $4$ , thus it must have the last two digits divisible by $4$ , which in our case is never true for $10$ is not divisible by $4$ .

Summing up: we have just proven that any $361\ldots 111$ (with two or more $1$ digits) can't be the square of an integer number $k$ (see first line), so the answer is that there are no other $1444\ldots 444$ that are perfect squares.

Let 144...4 (with n fours) be a perfect square. Since its even, we may let 4k^2 = 144...4=10^n +44...4. This implies, k^2 = 25×10^(n-2) + 11...1. Now, if n is greater than 3, then 25×10^(n-2) is a multiple of 4 and the remaining 111....1=11...100+11, which is of the form 4a+3. So the whole RHS is of the form 4b+3, which can never be a perfect square.

If $x^2=14...4$ then $x^2-4=(x+2)(x-2)$ is a multiple of 10, so x is even and $x=+/-2(mod 5)$ : combined the last digit of x must be 2 or 8. Below, a, b and c are nonnegative integers.

1) Suppose $x=10a+2$ . Then $x^2 = 100a^2+40a+4$ $\implies$ a must end on 1 or 6 $\implies$ x must end on 12 or 62

1.1) Suppose $x =100b+12$ . Then $x^2 = 10000b^2+2400b+144$ , implying that the 3rd digit from the right in $x^2$ is odd. The only possibility satisfying the pattern is $x^2 = 144$ .

1.2) Suppose $x =100b+62$ . Then $x^2 =10000b^2+12400b+3844$ . Now 4b+8 must end on 4, so b must end on a 4 or a 9, so x must end on 462 or 962.

1.2.1) Suppose $x=1000c+462$ . Then $x^2=1000000c^2+2×1000×462×c+213444$

1.2.2) Suppose $x=1000c+962$ . Then $x^2=1000000c^2+2×1000×962×c+925444$

For both these cases there is no solution because the digit at position 4 from the right is odd (and it cannot be the leading 1 since $x^2 > 1444$ ).

Summarizing: The only solution for possibility 1) is $x=12$ .

2) Suppose x ends on an 8, so $x=10a+8$ . Then $x^2 = 100a^2+160a+64$ $\implies$ 6a+6 must end on a 4 $\implies$ 6a ends on an 8 $\implies$ a ends on a 3 or 8. $\implies$ x ends on 38 or 88

2.1) suppose that x ends on 38. $x^2=(100b+38)^2=10000b^2+7600b+1444$ . Possibilities are: $b=0$ or $b=5 (mod 10)$ $\implies$ x ends on 038 or 538

2.1.1) if x ends on 038, the 4th position from the right is odd so $b=0$ $\implies$ $x=38$ is the only solution.

2.1.2) if x ends on 538, $x^2=(1000c+538)^2=1000000c^2+1076c+289444$ . Here too, position 4 from the right is odd and $x^2 >1444$ .

2.2) suppose that x ends on 88. $x^2=(100b+88)^2=10000b^2+17600b+7744$ . Position 3 from the right is odd and $x^2 > 144$ .

Summarizing: The only solution for possibility 2) is x=38, and 12 and 38 are the only solutions for x.

I was lazy and wrote a computer program. I tested it to 100,000,000 numbers and decided that I was pretty sure the answer was no.

        function checkNumber (n) {
        var arg = n.toString();
        var arr = arg.split("");

        if (arr[0] === "1") {
            var counter = 0;
            for (var i = 1; i < arr.length; i++) {
                if (arr[i] === "4") {
                    counter++;
                }
            } 

            if (counter === arr.length - 1) {
                return true 
            } else { 
                return false
            }

        } else {
            return false;
        }
    }

    var check = 0;
    for (var i = 4; i <= 100000000; i++) {
        if (checkNumber(i*i)) {
            check++;
        }
    }
    console.log(check)

python for i in range(10000): for j in range(200): if i*i==int('1'+'4'*j): print(i,flush=1) I used brute force.