State whether the following statement is true or false:
The number 7928 is a perfect square.

Hint: We will first look at the table of squares from 0 to 9. Then compare its last digit with the last digit of the given number 7928. If they match, 7928 might be a square, then we will have to do the prime factorization. Otherwise, it is definitely not a perfect square.

Complete step-by-step answer:
Let’s look at the square of the numbers from 0 to 9:-
\[0{\;^2} = 0\]
\[{1^2} = 1\]
${2^2} = 4$
${3^2} = 9$
${4^2} = 16$
${5^2} = 25$
${6^2} = 36$
${7^2} = 49$
${8^2} = 64$
${9^2} = 81$
We can clearly observe that there is no digit from 0 to 9 which has the last digit as 8 which is the last digit of 7928.
Hence, it cannot be a perfect square.
Hence, the statement is false.

Note: We must read the question properly as there might be cases that we are already given no in the question. For example:- The number 7928 is not a perfect square. Then our answer will be true.
We should observe that if some digit’s square’s last digit from 0 to 9 matches the last digit of some number, then we cannot conclude that it will definitely be a perfect square. This is just a necessary condition, not a sufficient one. So, if the digits do match, then we will have to find the prime factorization of the number and club the identical factors to form a pair. If all the factors get clubbed, we have a perfect square and if they do not, then it is not a perfect square.
Fun Fact:- The 4th of April 2016 is a Square Root Day, because the date looks like 4/4/16.
The next after that is the 5th of May 2025 (5/5/25).
Square of an even number is always even because if $n$ is even, then $n = 2m$ for some m in integers.
Then ${n^2} = {(2m)^2} = 4{m^2} = 2(2{m^2})$ which is even.
Squares of odd numbers are always odd because if $n$ is odd, then $n = 2m + 1$ for some m in integers.
Then ${n^2} = {\left( {2m + 1} \right)^2} = 4{m^2} + 1 + 4m = 2(2{m^2} + 2m) + 1$, which is odd.