A natural number $a$ has four digits such that $a^2$ ends with the exactly same four digits as that of $a$. Find the value of $a$?
The question was asked in a subjective maths olympiad and is certainly not missing any detail. The answer is 9376 as the square of 9376=87909376 but i am not able to understand how to solve it. I got it thanks ajotatxe
You must solve $$a^2-a=a(a-1)\equiv0\pmod{10,000}$$
Note that $a$ or $a-1$ must be multiple of $5^4=625$.