Question

How many four-digit numbers are there with the property that it is a square and the number formed by increasing all its digits by 1 is also a square.
[a] 0
[b] 1
[c] 2
[d] 4

Answer 1

Hint: Assume that the number is “abcd”. Hence find the number obtained by increasing each of the digits by 1. Use the fact that both the numbers are perfect squares to arrive at an equation. Check all the possible integral solutions of the equation. Hence find the total number of number satisfying the above condition

Complete step-by-step answer:
Let the number be “abcd” where a is the digit at thousands place, b is the digit at hundreds place, c is the digit at tens place, and d is the digit at units place.
Hence, we have
Original Number = 1000a+100b+10c+d
Hence the new number = 1000(a+1)+100(b+1)+10(c+1)+d+1=1000a+100b+10c+d+1111
Since both the numbers are perfect squares, we have
$\begin{aligned} & 1000a+100b+10c+d={{k}^{2}}\text{ -----(i)} \\\ & 1000a+100b+10c+d+1111={{j}^{2}}\text{ ------ (ii)} \\\ \end{aligned}$
Subtracting equation (i) and equation (ii), we get
$1111=\left( j+k \right)\left( j-k \right)$
Hence j+k and j-k are the factors of 1111
Now, we know that $1111=1111\times 1=101\times 11$
Hence, we have
J+k = 1111 and j-k = 1 or j+k = 101 and j-k = 11
For the first system we have
J+k = 1111
j-k = 1
Adding , we get
2j = 1112
i.e. j = 556
Hence k =555
Now ${{556}^{2}}=309136$, which is a five digit number.
Hence the above system has no solution
For the second system, we have
j+k = 101
j-k =11
Adding, we get
2j = 112
i.e. j = 56
Hence k = 45.
Now ${{56}^{2}}=3136$ and ${{45}^{2}}=2025$
Also, every digit of 3136 is one more than the corresponding digit in 2025.
Hence the four numbers such that it is a perfect square and the number formed by increasing each digit of the number by 1 is also a perfect square is 2025.
Hence, only one such four-digit number exists.
Hence option [b] is correct.

Note: [1] We have taken the larger of the two factors as j+k and smaller of the two factors as j-k , since j>0 and k>0 and hence j+k>j-k.
[2] The solution of the system of the linear equations formed does not guarantee the existence of the solution. This is why we checked whether all the conditions are satisfied to remove extraneous roots.