# How to solve this number theory question from INMO?

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?

• Using the exact same reasoning as in the previous paragraph. – Peter Taylor Dec 25 '18 at 7:21

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 $$3\cdot 7^2<7^3$$.