The Digits ABCABD

Note that we can re-write $N$ as $N=1001(100a+10b+c)+1$ . Let $N=n^2$ . We will look for solutions that work for n using modular arithmetic. Taking the equation mod 7, we get $n^2=1 \pmod{7}$ . Similarly we also have $n^2=1 \pmod{11}$ and $n^2=1 \pmod{13}$ . So we have a few cases to try. (Although there appears to be a lot of cases, several of them follow from others.):

1) $n=1 \pmod {7,11,13}$ . In this case we see that there are no values of n that work where a>0, since $n=1$ works and the Chinese Remainder Theorem tells us that this is the only solution (mod 7 11 13=1001). Hence there are no solutions in this case.

2) $n=1$ (mod 7,11) and $n=-1$ (mod 13). We quickly see that n=155. However, since $155^2$ has less than 6 digits, we see that this case fails.

3) $n=1$ (mod 7,13) and $n=-1$ (mod 11). We can see that n=274. But similarly, $274^2$ has less than 6 digits, and so this case fails too.

4) $n=1$ (mod 11,13) and $n=-1$ (mod 7). We see that n=573, and since $573^2$ has 6 digits this number works.

5) $n=-1$ (mod 7,11) and $n=1$ (mod 13) Note that this is analogous to case 2, except for the fact that the sign of $n$ is changed. Thus we quickly see that $n=-155=846$ (mod 1001) works.

6) $n=-1$ (mod 7,13) and $n=1$ (mod 11). Similarly, we have $n=-274 =727$ (mod 1001).

7) $n=-1$ (mod 11,13) and $n=1$ (mod 7). Similarly, we have $n=-573 =428$ (mod 1001).

8) $n=-1$ (mod 7,11,13) This simply yields $n=1000$ (mod 1001), which fails. Therefore, there are 4 total possible solutions.

If $N = \overline{abcabd} = M^2$ , then $M^2-1 = 1001\times\overline{abc}$ , and hence $M^2-1$ is divisible by $1001 = 7\times11\times13$ . Since $100000 \le N < 1000000$ , we deduce that $316 < M < 1000$ . There are eight cases to consider (omitting the details of the Chinese Remainder Theorem calculations):

1. If $M \equiv 1$ modulo $7,11,13$ , then $M \equiv 1$ modulo $1001$ , and no such $M$ exists in the desired range.

2. If $M \equiv 1$ modulo $7,11$ , and $M \equiv -1$ modulo $13$ , then $M \equiv 155$ modulo $1001$ , and so no $M$ exists in the desired range.

3. If $M \equiv 1$ modulo $7,13$ and $M \equiv -1$ modulo $11$ , then $M \equiv 274$ modulo $1001$ , and so no $M$ exists in the desired range.

4. If $M \equiv 1$ modulo $11,13$ and $M \equiv -1$ modulo $7$ then $M \equiv 573$ modulo $1001$ , so that $M=573$ and $N = 328329$ .

5. If $M \equiv 1$ modulo $7$ and $M \equiv -1$ modulo $11,13$ , then $M \equiv -573$ modulo $1001$ , so that $M=428$ and $N = 183184$ .

6. If $M \equiv 1$ modulo $11$ and $M \equiv -1$ modulo $7,13$ , then $M \equiv -274$ modulo $1001$ , so that $M = 727$ and $N=528529$ .

7. If $M \equiv 1$ modulo $13$ and $M \equiv -1$ modulo $7,11$ , then $M \equiv -155$ modulo $1001$ , so that $M=846$ and $N=715716$ .

8. If $M \equiv -1$ modulo $7,11,13$ then $M \equiv -1$ modulo $1001$ , and so no $M$ exists in the desired range.

Thus there are just $4$ numbers $N$ of the desired type.

$N = \overline{abcabd} = \overline{abcabc} + 1 = \overline{abc} (1001) + 1$ .

Let $N = n^{2}$ , where $n$ is a positive integer. Thus $\overline{abc} (1001) + 1 = n^{2}$ or $\overline{abc} (1001) = n^{2} -1 = (n-1)(n+1)$ .

As $100 \leq \overline{abc} \leq 999$ , so $100(1001) + 1 \leq \overline{abc} (1001) + 1 \leq 999(1001) + 1$ .

Thus $316^{2} = 99856 \leq 100101 \leq n^{2} \leq 1000^{2} - 1 + 1 = 1000^{2}$

As $n > 0,316 \leq n \leq 1000$ .

Consider $\overline{abc} (1001) = (n-1)(n+1)$

The prime factorisation of $1001 = 7 * 11 * 13$ . As $n-1$ and $n+1$ have difference 2, they cannot have common factor larger than 2. The equation above shows that the prime factors 7, 11 and 13 divide $n-1$ or/and $n+1$ .So 7, 11, 13 each divide only $n-1$ or $n+1$ .

It can be seen that 2 of the primes divide 1 of $n-1$ and $n+1$ while the last prime divides the other.

There are 3 cases. First, $n-1$ or $n+1$ is divisible by $11 * 13 = 143$ while the other is divisible by 7. Secondly, $n-1$ or $n+1$ is divisible by $7 * 13 = 91$ while the other is divisible by 11. Lastly, $n-1$ or $n+1$ is divisible by $7 * 11 = 77$ while the other is divisible by 13.

We have $315 \leq n-1 < n+1 \leq 1001$ . Thus we only need check the multiples of 143, 91 and 77 in this range and set them as $n-1$ or $n+1$ while checking whether the other of the pair passes the test of divisibility by the 3rd prime.

This yields $(n-1,n+1) = (427,429), (572,574), (726,728), (845,847)$ .

So $n = 428, 573, 727, 846$ . From $\overline{abc} (1001) = n^{2} -1$ , we obtain
$\overline{abc} = \frac{n^{2} -1}{1001}$ . So $\overline{abc} = 183, 328, 528, 715.$

Therefore $N = \overline{abcabd} = 183184, 328329, 528529, 715716$ .

Based on my finding, I used up an hour to find the squared of from $317$ to $999$ . Since $\sqrt {100 000} = 316.227766016$ . Among all $683$ numbers, there are $6$ perfect squares satisfy the following condition of $abcabd$ . These numbers are...

$428^{2} = 183 184$ $548^{2} = 300 304$ $573^{2} = 328 329$ $727^{2} = 528 529$ $846^{2} = 715 716$ $856^{2} = 732 736$

Since $d = c + 1$ , $300 304$ and $732 736$ does not satisfy the following condition. Hence, there are $\boxed{4}$ solutions.

Let $n:=\overline{abc}$ . With $d=c+1$ , we can simplify the perfect square $N$ : $\begin{aligned} N&=\overline{abcabc}+1=1001n+1\overset{!}{=}m^2 &&&\Rightarrow&&&& 1001n&=(m+1)(m-1)&&&&&\left| 1001=7\cdot 11\cdot 13 \right. \end{aligned}$

Each of the prime factors of 1001 must divide one of the two factors on the right. If we collect the prime factors into $p,\: q$ , respectively, we get a different diophantine equation to solve for each factorization of 1001: $\begin{aligned} 1001&=pq: &&& \left.\begin{aligned} m+1&=px\\ m-1&=qy \end{aligned}\right\}&& 2&=px-qy,& x,\:y&\in\mathbb{Z}&&&&&\left|m^2-1=pqxy,\right.&&n&=\frac{m^2-1}{1001}=xy \end{aligned}$

Notice the value of $m^2-1$ does not change if we switch the signs of both $p,\:q$ , or if we interchange them: It is enough to consider factorizations of 1001 with $0<p\leq q$ . We are left with only four cases, and $p,\:q$ are coprime in all of them:

$\begin{aligned} \begin{array}{r|r|l c l l | l} p & q & &\red{(*)} & & &n=xy\\[.25em] \hline 1 & 1001 & 2=x-1001y& \:\: \Rightarrow\:\: & \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 1&1001\\ 0&1 \end{pmatrix}\begin{pmatrix}2\\k\end{pmatrix}, &\:\: k\in\mathbb{Z}& (2+1001k)k\\[1em] 7 & 143 & 2=7x-143y& \Rightarrow & \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 41&143\\ 2&7 \end{pmatrix}\begin{pmatrix}2\\k\end{pmatrix}, &\:\: k\in\mathbb{Z}& (82+143k)(4+7k)\\[1em] 11 & 91 & 2=11x-91y& \Rightarrow & \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 25&91\\ 3&11 \end{pmatrix}\begin{pmatrix}\red{1}\\k\end{pmatrix}, &\:\: k\in\mathbb{Z}& (25+91k)(3+11k)\\[1em] 13 & 77 & 2=13x-77y& \:\: \Rightarrow\:\: & \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 6&77\\ 1&13 \end{pmatrix}\begin{pmatrix}2\\k\end{pmatrix}, &\:\: k\in\mathbb{Z}& (12+77k)(2+13k) \end{array} \end{aligned}$ Our unknown number is restricted by $n\in\{102;\:\ldots;\:987\}$ . These restrictions determine which $k$ we may choose. $\begin{aligned} \begin{array}{r|r|r|r} p & k & n & N \\[.5em]\hline 1 & & & \\[.5em]\hline 7 & -1& 183& 183184\\[.5em] & 0 & 328 & 328329\\[.5em]\hline 11 & -1& 528&528529\\[.5em]\hline 13 & -1 & 715&715716 \end{array}&&&&\Rightarrow &&&&\boxed{4}\text{ distinct solutions!} \end{aligned}$ It's easy to see there are no other possibilities, because the parabolas $xy$ all have both zeros and their global minimum in $[-1;\:0]$ .

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Rem.: In $\red{(*)}$ , we use the solution for the general linear diophantine equation: $\begin{aligned} px-qy&=r,&&&p,\:q,\:r&\in\mathbb{Z},& \gcd(p,\:q)&=1 &&&\Rightarrow &&&&\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix} x_0 & q\\ y_0 & p \end{pmatrix}\begin{pmatrix}r\\k\end{pmatrix},&k\in\mathbb{Z} \end{aligned}$

The pair $(x_0,\:y_0)$ is any solution of $px_0-qy_0=1$ , obtained by Euclid's Algorithm or simply guessing. In the third row of the table, $k$ was shifted to get the smallest non-negative solution at $k=0$ .

$N=100100a+10010b+1001c+1$ . Let $N=x^2$ so that $316<x<1000$ and $N-1=1001(100 a + 10 b + c)=(x+1)(x-1)$ . Since $1001=7×11×13$ we have

$7×11×13×(100 a + 10 b + c)=(x+1)(x-1)$

Each prime factor must be a factor of either $x+1$ or $x-1$ . So, respecting the upper bound for x we cannot have them all in either $x+1$ or $x-1$ . There are six possible cases

• x+1 a multiple of 7×11 and  x-1 a multiple of 13. Or $x \equiv -1 \pmod{77}$ and $x \equiv 1 \pmod{13}$ we find $x=846 \pmod {1001}$ . $N=715716$
• x+1 a multiple of 13 and  x-1 a multiple of 7×11, Or $x \equiv 1 \pmod{77}$ and $x \equiv -1 \pmod{13}$ we find $x=-846=155 \pmod {1001}$ . $N=24025$ . This disqualifies as a=0.
• x+1 a multiple of 7×13 and  x-1 a multiple of 11. $x \equiv -1 \pmod{91}$ and $x \equiv 1 \pmod{11}$ we find $x=727 \pmod {1001}$ . $N=528529$
• x+1 a multiple of 11 and  x-1 a multiple of 7×13. we find $x=-727=274 \pmod {1001}$ . $N=75076$ This disqualifies as a=0.
• $x \equiv -1 \pmod{143}$ and $x \equiv 1 \pmod{7}$ , $x=428 \pmod {1001}$ . $N=183184$
• $x \equiv 1 \pmod{143}$ and $x \equiv -1 \pmod{7}$ , $x=-428=573 \pmod {1001}$ . $N=328329$

So we have $\boxed{4}$ solutions.

As an example for finding x I show the 3rd case below

$x\equiv-1\pmod{91}$

$x=90+91t$

$x\equiv2+3t \pmod{11}$

Also, $x\equiv1 \pmod{11}$

$3t\equiv-1\equiv-12 \pmod{11}$

$t\equiv-4\equiv7 \pmod{11}$

$x=90+91×7=727$

$N=x^2=528529$

Use PARI.