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I need to find the integers that when squared they maintain the las two digits, I've started like this: being b and a the last two digits of the number $(10b+a)²\equiv 10b+a \mod(100) \Rightarrow 20ab + a² \equiv 10b+a \mod(100) $ I don't know how to continue this, any idea?

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  • $\begingroup$ Hint: how many solutions are there to $n^2\equiv n\pmod {10}$? if $n$ is one of those...can you lift it to a solution of $n^2\equiv n\pmod {100}$ $\endgroup$ – lulu May 21 '16 at 19:38
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HINT: Start by checking which digits $a$ can be; there are only four possibilities. Then substitute each of those possibilities into your congruence

$$20ab+a^2\equiv 10b+a\pmod{100}$$

and solve for $b$; note that there may be no solution in some cases.

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  • $\begingroup$ Okey, I found 00, 01, 25 and 76, thank you. $\endgroup$ – joan capell May 21 '16 at 19:56
  • $\begingroup$ @joan: Looks good; you’re welcome! $\endgroup$ – Brian M. Scott May 21 '16 at 20:01

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