Prove that x^2+17xy+17y^2+1 ≡ 0 (mod m) for any m.

Makuta

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Prove that x^2+17xy+17y^2+1 ≡ 0 (mod m) for any m.

I found that it has roots in rational numbers and if we put those roots in the comparison it'll be correct, because eventually we get (-1+1) ≡ 0 (mod m). How can I show that it is true for any m, using that information? Is there a helpful theorem? I stuck and do not see what's the next step should be.
 
Prove that x^2+17xy+17y^2+1 ≡ 0 (mod m) for any m.

I found that it has roots in rational numbers and if we put those roots in the comparison it'll be correct, because eventually we get (-1+1) ≡ 0 (mod m). How can I show that it is true for any m, using that information? Is there a helpful theorem? I stuck and do not see what's the next step should be.
I must be missing something here. Are you saying that this is supposed to be identically true for any integer m, meaning for any integer pair (x,y)? Clearly for x=0. y=0, the LHS is 1, not zero, modulo any m!

Please show an image of the original problem.
 
I must be missing something here. Are you saying that this is supposed to be identically true for any integer m, meaning for any integer pair (x,y)? Clearly for x=0. y=0, the LHS is 1, not zero, modulo any m!

Please show an image of the original problem.
Here, sorry for confusion:1717002216500.png
 
Ah! That does change things.

I think "comparison" here means what I would call "congruence", and "module" is what I would call a "modulus"; showing that it's solvable, I assume means that the congruence is true for some pair (x,y) of integers, rather than for all.

Now we can start thinking about it.
So how can I continue proving this?
 
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