Hey thanks Tonio

Though hmm now is there a chance I could be right?

On the top of pg 16 on here:

http://www.warwick.ac.uk/~maseap/tea...t/ntnotes1.pdf
the Chinese Remainder Theorem is outlined.

So I had got y is congruent to 87 mod 125

y is congruent to 1 mod 4

Then I wrote that 125 = (31x4) + 1.

4= (4x1) + 0

Therefore gcd(125, 4) = 1.

1 = 5^3 - (31 x 4)

So let u = 5^4

v = -31x4

Then u is congruent to 1 mod 4 and 0 mod 5

And v is congruent to 1 mod5 and 0 mod 4.

Therefore the values of y solving the simultaneous congruences 87 mod 125

and 1 mod 4 are:

y congruent to (87 x (-31x4)) + (1x 5^4) [mod (5^3x4)].

That was the answer I wrote down as my final answer because of the top of page 16.

I really have not much idea what you've done above (based on your OP...I sure understand what you're doing ), and this is one of the worst things that can happen to one of my students (which you are not...fortunately!), since if I don't understand something I dismiss it and that student's grade goes on holiday to Bahamas... It seems to be that you're saying that $\displaystyle y=(87 \cdot (-31\cdot4)) + (1\cdot 5^4)=-10,163$ is a solution to the above system, and it indeed is (what does the "mod (5^3x4)" means, anyway? The solution must be an** integer**, not some integer modulo something!) What you probably meant is that the solution must be any integer satisfying $\displaystyle y=-10,163\!\!\!\pmod{125\cdot 4=500}$ , which of course is correct, and then you can take, for example, the solution $\displaystyle y=-10,163+21\cdot 500=337$ , which is, imo, nicer and positive. So...? It was **you** that said that $\displaystyle y=87$ is a solution, and I didn't check your work, so based in what you said I told you that it can't be since $\displaystyle 87\neq 1\!\!\!\pmod 4$ ...After that I assumed you checked all the other possibilities and since 97, 107 and 117 didn't work then there's no solution. Certainly , since $\displaystyle (125,4)=1$ one can try the CRT...but you didn't even mention the CRT until your last messages! You only mentioned Hensel's lemma, which is a nice, powerful, tool,and what you did above is the algorithm to use based in Euclides' algorithm... Good, so you found the general solution to the system $\displaystyle y^3=3\!\!\!\pmod{125}\,,\,y=1\!\!\!\pmod 4$ , and you must have written that it is any integer of the form $\displaystyle y=-10,163=337\!\!\!\pmod{500}$ Tonio
Sorry to bring this up again - but surely this must work. Since the question specifically asked:

'Solve the following system of congruences: y^3 is congruent to 3 (mod 5^3) and y is congruent to 1 (mod 4)'

which I think is the exact same question as on top of pg 16 except first we have to use Hensel's lemma to reduce y^3 congruent to (mod p^k) to y congruent to (mod p^k) and then surely we can apply top of pg 16???

Sorry Tonio for not putting this down earlier. I would be really grateful if you could check. Thanks very much.