METHOD 1
If the two perfect squares are m2 and n2, then:
m2-n2=(m-n)(m+n)=3333=3×11×101.
So 33×101=303×11=3×1111.
Taking each factor pair in turn:
m-n=33, m+n=101; 2m=134, m=67 and n=34;
m-n=11, m+n=303; 2m=314, m=157 and n=146;
m-n=3, m+n=1111; 2m=1114, m=557 and n=554.
The only answer that fits the requirements of 2 4-digit numbers is m=67 and n=34. m2=4489, n2=1156, 4489-1156=3333.
The method below is a totally different (and lengthier) logical approach.
METHOD 2
Let 1000a+100b+10c+d=n2 and 1000(a+3)+100(b+3)+10(c+3)+d+3=m2.
Because each digit cannot exceed 9, then we know that a, b, c, d must be 6 or less.
If we subtract the smaller from the larger we get:
m2-n2=3333.
Perfect squares end with 0, 1, 4, 9, 6 or 5 so the differences between two perfect squares are limited as shown in the table below:
|
0 |
1 |
4 |
9 |
6 |
5 |
| 0 |
0 |
9 |
6 |
1 |
4 |
5 |
| 1 |
1 |
0 |
7 |
2 |
5 |
6 |
| 4 |
4 |
3 |
0 |
5 |
8 |
9 |
| 9 |
9 |
8 |
5 |
0 |
3 |
4 |
| 6 |
6 |
5 |
2 |
7 |
0 |
1 |
| 5 |
5 |
4 |
1 |
6 |
9 |
0 |
Subtract the digit in the top row from each of the digits in the left column, adding 10 if the result is negative.
Consider the two cases where the larger number ends in 4 and the smaller in 1; or the larger number ends in 9 and the smaller in 6 (red 3). Squares ending in 4 have 2 or 8 as the last digit of the square root; squares ending in 1 have 1 or 9 as the last digit of the square root. Squares ending in 9 have 3 or 7 as the last digit of the square root; squares ending in 6 have 4 or 6 as the last digit of the square root.
Also, 4-digit squares have 2-digit square roots.
The first digit a must be between 1 and 6 (so the larger number must start with a number between 4 and 9).
If we start with the 2-digit square roots, we need to test only a few numbers, starting with square roots ending in 7 or 3 (larger numbers), which give us squares ending in 9, or square roots ending in 4 or 6 (smaller numbers), which give us squares ending in 6. We can start with the smaller square root numbers between 32 and 99: 34, 36, 44, 46, 54, 56, 64, 66, 74, 76, 84, 86, 94, 96. (322=1024, the smallest number with a 4-digit square. 992=9801, the largest number with a 4-digit square.)
Square these numbers: 1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776. There's no point continuing this sequence because the leading digit > 6.
I've crossed out those containing digits greater than 6.
Now add 3333: 4489, 5449, 6469, 7689. Of these only 4489 is a perfect square=672. So the pair is 1156 and 4489. Now we switch to the other set of numbers for further investigation. Applying the same technique, we start with the list: 39, 41, 49, 51, ..., 79, 81. Squaring these we get a set of 6 valid squares (removing any containing digits > 6). Adding 3333 to each of these we end up with no perfect squares.
So it would appear that the only solution is the pair 1156, 4489 (342, 672).