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I am trying to show that if $p \equiv 2 \pmod{3}$, then for any $a \in \mathbb{Z}$ the congruence equation $$ x^3 \equiv a^2 - 7 \pmod{p} $$ has a solution.

I was thinking that if the equation $x^3 - a^2 + 7 = 0$ does not have any solutions modulo $p$, then it is irreducible over $\mathbb{Z}_p$ and hence must be an irreducible factor of the polynomial $x^{p^3} - x$ over $\mathbb{Z}_p$. It seems that comparing coefficients of $x$ might then lead to a contradiction but I haven't been successful in doing this.

Any hints or suggestions as to how to proceed will be highly appreciated.

MAZ
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    FYI, using an Approach0 search, there's the somewhat related Is $y^2=x^3+7$ unsolvable modulo some $n$?. Although this shows there's always a solution for any modulo, not just with your restriction, it doesn't show that there's always a solution for every possible value of $a$. Nonetheless, I thought you might find it interesting, and possibly also somewhat useful. – John Omielan Apr 16 '25 at 18:34
  • Please delete this question since we already have too many copies of these arguments, which makes it difficult to locate the best answers by search. You will need to unaccept the answer before deleting (click the checkmark), and maybe wait for the answr to be deleted. – Bill Dubuque Apr 16 '25 at 19:42
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    You can solve $$x^3\equiv b\pmod p$$ for any $b$ if $p\equiv 2\pmod 3.$ The closed form is $x=b^{(2p-1)/3}.$ – Thomas Andrews Apr 16 '25 at 23:27

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The multiplicative group $\mathbb{Z}/p\mathbb{Z}^\times$ is a cyclic group of order $p-1$. So the map

$$x \mapsto x^3$$
is bijective when $\gcd(3, p-1) = 1$, which implies there exist $x$ for any $x^3$.

aerile
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  • I was considering this map but I don't know why I didn't think of it as a group homomorphism and use the cyclicity of $(\mathbb{Z}/p\mathbb{Z})^{\times}$. I was trying to show that this map is injective which is painful to do without the above apparatus. Thanks. – MAZ Apr 16 '25 at 19:35
  • Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. site policy announcement here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe/PSQ processing.) – Bill Dubuque Apr 16 '25 at 19:41
  • The linked posts only state the general fact; my answer shows it can be applied to this specific problem, so it isn’t a duplicate. – aerile Apr 22 '25 at 17:48