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Solve the following equation in integers:

$$m(4m^2 + m + 12) = 3(p^n -1)$$

where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.

This simplifies to :

$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$

$\gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.

Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7\text{ or }49$$

This means the prime $p$ is 7. and $4m + 1 = 3*7^k\text{ or }7^k$

If $\gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $\gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 \leq 3*7^2 < 7^3 \leq 4m + 1$$

How do we arrive at the above inequality chain?

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  • 2
    $\begingroup$ Using the exact same reasoning as in the previous paragraph. $\endgroup$ – Peter Taylor Dec 25 '18 at 7:21
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Since $m^2+3 =7^l$ or $m^2+3= 3\cdot 7^l$ we see that $m^2+3 \leq 3\cdot 7^l$

Now we have $7^2\mid m^2+3$ and $7^3\not{\mid} \;m^2+3$ so $l=2$.

And since $7^3\mid 4m+1$ we have $7^3\leq 4m+1$. Clearly $3\cdot 7^2<7^3$.

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  • $\begingroup$ So, you don't like the answer? $\endgroup$ – Aqua Jan 4 at 21:44
  • $\begingroup$ Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks. $\endgroup$ – caffeinemachine Jan 26 at 16:26
  • $\begingroup$ There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest.... $\endgroup$ – Aqua Jan 26 at 21:52

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