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Who can prove that $7|4x^2+5x+2+d $ following solutions I found are correct (complete)? Where $d$ is integer constant (not relation with $x$), find $d$ which make $7|4x^2+5x+2+d $ have solution, and give corresponding solution $x$ as well.

I have found following would be correct solutions, but not sure if them are complete: $d≡0~ mod~7~⟹~ (x=2+7t)$; $d≡3~ mod~7~⟹~ (x=1+7t, x=3+7t)$; $d≡5~ mod~7~⟹~ (x=4+7t, x=7t)$; $d≡6~ mod~7~⟹~ (x=5+7t, x=6+7t)$.

My approach is find them by testing $d$ value one by one by, then I can get that $d$ with pattern $d≡{0,3, 5, 6}~ mod ~7$ and the copressponding $x$ can be sloved, but I couldn't prove that these are complete solutions.

A more genral question is when $ pq|ax^2+bx+c$ has solution, where $pq$ is semiprime, $a,b,c$ are integer constant, find $x$. If not easy to answer, do we have a general algorithm within limit steps to detect any given $pq, a,b,c$ for $ pq|ax^2+bx+c $ has solution or has no solution?

xMath
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    Any $x$, and $d=6(4x^2+5x+2)$, is a solution. – David Mar 10 '23 at 06:13
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    Sorry, not solve d, it is a constant. What is form of $x$? – xMath Mar 10 '23 at 06:21
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    All you have to do is check what $d$ you need for $x=0,1,2,\dots,6$, and that gives you all the information you need (for the question in the title). – Gerry Myerson Mar 10 '23 at 06:48
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    For the general question, you want to check whether or not $b^2-4ac$ is a quadratic residue modulo $p$ and modulo $q$, and for that you want to apply The Law of Quadratic Reciprocity, so now you know what to look up. – Gerry Myerson Mar 10 '23 at 06:50
  • What did you try? Where were you stuck? https://math.stackexchange.com/help/how-to-ask – Anne Bauval Mar 10 '23 at 07:31
  • For $d\in{0,3,5,6} \bmod 7$ (and those are all integers), the original equation has infinitely many integer solutions, so I'm puzzled by why you think anyone can prove that there are none? – Henrik supports the community Mar 10 '23 at 10:05
  • @Henrik, I think what OP is asking is, for which values of $d$ is there a solution, and for which values of $d$ is there no solution. – Gerry Myerson Mar 10 '23 at 11:37
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    @Anne Bauval, I found when $d∈{0,3,5,6} mod 7$ I have solution for $7|4x2+5x+2+d$, but I can't prove such. The solutions are: $d≡0 mod 7 (x=2+14t,x=9+14t); d≡3 mod 7 (x=1+7t,x=3+5t); d≡5 mod 7 (x=4+5t,x=7t); d≡6 mod 7 (x=5+7t,x=6+7t), where t∈Z$. – xMath Mar 12 '23 at 13:42
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    When $d\equiv0\bmod7,$ simply write $x=2+7t.$ When $d\equiv3,$ misprint (replace $+5t$ by $+7t$). When $d\equiv7$ or $4,$ the results in your post are wrong (they should be the same as when $d\equiv0$). Anyway, you can suppress from your post all the cases you considered after $d\equiv6$ (they are redundant). As for your more general question, Gerry already answered you in a comment, but imo you should have asked that in a distinct post. – Anne Bauval Mar 12 '23 at 16:32
  • @AnneBauval, thanks so much, as a summary, following would be correct solutions: $d≡0 mod7⟹ (x=2+7t)$; $d≡3 mod7⟹ (x=1+7t, x=3+7t)$; $d≡5 mod7⟹ (x=4+7t, x=7t)$; $d≡6 mod7⟹ (x=5+7t, x=6+7t)$. – xMath Mar 13 '23 at 03:07
  • Sure, and the proof is easy if you follow Gerry's advice above: "All you have to do is check what $d$ you need for $x=0,1,2,…,6$". E.g. for $x\equiv5\bmod7,$ you find ($\bmod7$) $4x^2+5x+2+d\equiv4(-2)^2+5(-2)+2+d\equiv1+d\equiv-6+d,$ which is divisible by $7$ iff $d\equiv6\bmod7.$ – Anne Bauval Mar 13 '23 at 08:19
  • @AnneBauval. I got it, many thanks for you and Gerry. – xMath Mar 13 '23 at 10:35

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