<s>[INST] Find $x$ such that $x \equiv7\pmod {37}$ and $x^2 \equiv 12\pmod {37^2}$ Find $x$ such that $x \equiv7 \pmod {37}$ and $x^2 \equiv 12\pmod {37^2})$
My attempt: Given $x \equiv7\pmod {37}$ 
so $37|(x-7)$ so $37^2|(x-7)^2$
so $x^2-14x+49 \equiv 0\pmod {37^2}$
as $12-14x+49 \equiv 0\pmod{37^2}$ [as $x^2 \equiv 12\pmod {37^2}$]
so $14x\equiv 61\pmod {37^2}$  now I can find a set of solution using Euclidean algorithm 
I would like some one to verify if this is correct as it seems longer and also is there a better and shorter method 
 [/INST] I would do $$(37n+7)^2=12\pmod{37^2}\\14n*37+49=12\pmod{37^2}\\14n+1=0\pmod{37}$$
which has a smaller modulo
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