$N$ be a natural number consisting all digits except $0.$ If the last digit of $N$ is 5 then prove that $N$ can't be a perfect square. Let $N$ be a nine digited natural number consisting all digits except $0.$ If the last digit of $N$ is 5 then prove that $N$ can't be a perfect square.


I have have tried finding a prime $p$ dividing $N$ but $p^2$ not dividing $N$ but I am unable to do it that way.  Also I found that ten's digit of $N$ can be $2$ or $7.$

Any help will be appreciated.
Tip: the last digit of $N$ is $5$, so $\sqrt N$ must also end with $5$, but if that's the case, the tens digit of $N$ must be $2$, since no perfect square ends with $75$, but if that's the case...
Let P be that number. After dividing by 5, u get Q. Now Q cannot contain 5. So Q doesn't have 5 as a factor. So Q times 5 cannot be a square as a square should have prime factors of order 2.

Edit:
Using the following facts which can be proved individually.


perfect square ending in 5, will also end in 25.
perfect square ending in 25 will have either 0,2,6 in hundred's place.
perfect square ending in 625, will have 0,5 in thousand's place.


so the number N, for it to be perfect square, should end in 625( it cant have 2 or 0 in hundred's place). After that it cant have eihter 0 or 5 in thousand's place. so N is not a perfect square.