I wish to know, if it is possible to prove the following facts using any known facts (till elementary school level):
Only 1 and 9 are the perfect odd numbered squares which are entirely made up of odd numbers / digits.
There are NO other such perfect squares.
That means, if you take any other odd numbered square it would at least containg one even-numbered digit. For example: 25 / 49 / 81 etc.
Any clues / hints towards proof are also welcome !
This proof is adapted from a question from Dr Math.
Notice that the last two digits of $\color{red}{1^2}$ and $\color{blue}{99^2}$ ; $\color{red}{3^2}$ and $\color{blue}{97^2}$, $\color{red}{5^2}$ and $\color{blue}{95^2}$ and so on, are exactly the same. We can prove this using algebra; looking at the numbers $x^2$ and $x^2-200x+10000$, we see that only the $100$s place and $10000$s place are changed in the second number, but not the $1$s or the $10$s place.
We can do the exact same thing with $\color{red}{1^2}$ and $\color{blue}{49^2}$ ; $\color{red}{3^2}$ and $\color{blue}{47^2}$, $\color{red}{5^2}$ and $\color{blue}{45^2}$. With the two numbers $x^2$ and $x^2-100x+2500$, we can again see that the last two digits are not changed in the second number.
Now check the squares of the numbers from $1$ to $25$, and see which ones have an even digit. Since adding any multiple of $100$ to $x$ (shown by expanding $(x+100k)^2 = x^2+200xk+10000k^2)$ to these numbers will not affect the last two digits, your conclusion about these numbers holds regardless of the number's $100$s, $1000$s, $10000$s or higher places.