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Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$20^m - 10m^2 + 1 = 19^n.$
Source of the problem
South Africa MO 2019, Problem 6
Number Theory
7/10
Suggested Book
Excursion in Mathematics by Bhaskaracharya Prathisthan

Do you really need a hint? Try it first!

If you read this, you will get to know some techniques to explore Diophantine Equations. Let’s get an idea of m and n  by using the modulo technique. To get an idea of n, we must remove or eliminate m, to do that we take modulo 10. Observe that the equation demands to be taken modulo 10, given the numbers and it turns out that $19^n = (-1)^n = 1 mod 10$. It implies that n must be even. Try to get an idea of m now. Also, (0,0) is a solution. So, we take both m and n as non-zero.

To remove n, using the information that n is even, we can remove the variable n, taking modulo 4. So, $2m^2 + 1 = 1 mod 4$ implies m must be even. Let m = 2p.

Observe that RHS is a square and LHS $< 20^{2p}$.

The largest square $< 20^{2p}$ is $(20^{p} – 1)^2$. Thus,  $LHS \leq (20^{p} – 1)^2$. Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$,
which simplifies to   $p^2\ge20^{p-1}$. (*)
Now, this turns out to be inequality and this results in the solution p = 1. This gives the only solution (2,2) as (m,n).

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