Number Theory

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Formulas and Statements

(1.) The sum of the first n natural numbers: $\sum\limits_{k = 0}^{n} 1 + 2 + 3 + 4 + 5 + ... + n = \dfrac{n(n + 1)}{2}$

(2.) The sum of the first n squares $\sum\limits_{k = 0}^{n} 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$

(1.) Show that the sum of the first n squares $\sum\limits_{k = 0}^{n} 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6} \\[5ex]$ Use at least three different approaches/methods.

1st Approach: Method of Undetermined Coefficients.
This is based on these statements:
(1.) If the terms of a sequence is of degree n, then the sum of the series is of degree n + 1
Hence, the sum of the series of squares results in a cubic polynomial.

(2.) The difference between the sum of the terms of a series and the sum of the terms of the series less one term is the square of the terms of the series.
For example, the difference between the sum of n terms, say $S_n$ and the sum of $n - 1$ terms, say $S_{n - 1}$ is $n^2$

$ \underline{\text{Method of Undetermined Coefficients}} \\[3ex] \text{For } 1^2 + 2^2 + 3^2 + ... + n^2 \\[3ex] S_n = An^3 + Bn^2 + Cn + D...eqn.(1) \\[3ex] \text{where }A, B, C, D \text{ are coefficients} \\[3ex] S_{n - 1} = A(n - 1)^3 + B(n - 1)^2 + C(n - 1) + D \\[3ex] $ I want to use Pascal's Triangle for the expansion.
Use whatever approach you prefer. $$ \begin{array}{c} 1 \\[2ex] 1 \quad 1 \\[2ex] 1 \quad 2 \quad 1 \\[2ex] 1 \quad 3 \quad 3 \quad 1 \\[2ex] \end{array} $$ $ (n - 1)^3 \\[3ex] = n^3 + 3n^2(-1) + 3n(-1)^2 + (-1)^3 \\[3ex] = n^3 - 3n^2 + 3n - 1 \\[5ex] A(n - 1)^3 \\[3ex] = A(n^3 - 3n^2 + 3n - 1) \\[3ex] = An^3 - 3An^2 + 3An - A \\[5ex] (n - 1)^2 \\[3ex] = n^2 - 2n + 1 \\[5ex] B(n - 1)^2 \\[3ex] = B(n^2 - 2n + 1) \\[3ex] = Bn^2 - 2Bn + B \\[5ex] \implies \\[3ex] S_{n - 1} = An^3 - 3An^2 + 3An - A + Bn^2 - 2Bn + B + Cn - C + D \\[2ex] S_{n - 1} = An^3 + n^2(-3A + B) + n(3A - 2B + C) - A + B - C + D ...eqn.(2) \\[5ex] S_n - S_{n - 1} = n^2 \\[3ex] \underline{\text{LHS}} \\[3ex] eqn.(1) - eqn.(1) \implies S_n - S_{n - 1} \implies \\[3ex] An^3 - An^3 + n^2[B - (-3A + B)] + n[C - (3A - 2B + C)] + D - (-A + B - C + D) \\[3ex] n^2(B + 3A - B) + n(C - 3A + 2B - C) + D + A - B + C - D \\[3ex] n^2(3A) + n(-3A + 2B) + A - B + C \\[3ex] \underline{\text{RHS}} \\[3ex] n^2 \\[3ex] n^2 + 0n + 0 \\[3ex] \underline{\text{LHS = RHS}} \\[3ex] \text{Equate coefficients; find the unknown coefficients} \\[3ex] 3A = 1 \\[3ex] A = \dfrac{1}{3} \\[5ex] -3A + 2B = 0 \\[3ex] 2B = 3A \\[3ex] 2B = 3 * \dfrac{1}{3} \\[5ex] 2B = 1 \\[3ex] B = \dfrac{1}{2} \\[5ex] A - B + C = 0 \\[3ex] C = B - A \\[3ex] C = \dfrac{1}{2} - \dfrac{1}{3} \\[5ex] C = \dfrac{3 - 2}{6} \\[5ex] C = \dfrac{1}{6} \\[5ex] \text{We need to find } D \\[3ex] S_n = 1^2 + 2^2 + 3^2 + ... + n^2...\text{Given} \\[3ex] \text{From } eqn.(1) \\[3ex] S_n = An^3 + Bn^2 + Cn + D \\[3ex] \text{when } n = 1 \\[3ex] S_1 = 1^2 = 1 \\[3ex] \text{Also}: \\[3ex] S_1 = \dfrac{1}{3} * 1^3 + \dfrac{1}{2} * 1^2 + \dfrac{1}{6} * 1 + D \\[5ex] \implies \\[3ex] \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{6} + D = 1 \\[5ex] D = 1 - \dfrac{1}{3} - \dfrac{1}{2} - \dfrac{1}{6} \\[5ex] D = \dfrac{6 - 2 - 3 - 1}{6} \\[5ex] D = \dfrac{0}{6} \\[5ex] D = 0 \\[3ex] \implies \\[3ex] S_n = \dfrac{1}{3}n^3 + \dfrac{1}{2}n^2 + \dfrac{1}{6}n \\[5ex] S_n = \dfrac{n^3}{3} + \dfrac{n^2}{2} + \dfrac{n}{6} \\[5ex] = \dfrac{2n^3 + 3n^2 + n}{6} \\[5ex] = \dfrac{n(2n^2 + 3n + 1)}{6} \\[5ex] ................................................. \\[3ex] 2n^2 + 3n + 1 \\[3ex] 2n^2 + 2n + n + 1 \\[3ex] 2n(n + 1) + 1(n + 1) \\[3ex] (n + 1)(2n + 1) \\[3ex] ................................................. \\[3ex] \therefore S_n = \dfrac{n(n + 1)(2n + 1)}{6} \\[5ex] $ 2nd Approach: Proof by Mathematical Induction.

Steps	LHS	RHS
	$1^2 + 2^2 + 3^2 + ... + n^2$	$\dfrac{n(n + 1)(2n + 1)}{6}$
Initial Step Test for $n = 1$	First term: $1^2 = 1$	$ \dfrac{1(1 + 1)[2(1) + 1]}{6} \\[5ex] \dfrac{1(2)(3)}{6} \\[5ex] 1 $
The initial step works
Optional Testing Test for $n = 2$	First term + Second term: $ 1^2 + 2^2 \\[3ex] 1 + 4 \\[3ex] 5 $	$ \dfrac{2(2 + 1)[2(2) + 1]}{6} \\[5ex] \dfrac{2(3)(5)}{6} \\[5ex] 5 $
Optional Testing Test for $n = 3$	First term + Second term + Third term: $ 1^2 + 2^2 + 3^2 \\[3ex] 1 + 4 + 9 \\[3ex] 14 $	$ \dfrac{2(3 + 1)[2(3) + 1]}{6} \\[5ex] \dfrac{3(4)(7)}{6} \\[5ex] 14 $
The optional testing works
Induction Hypothesis Assume $n = k$ This is the $kth$ term $1^2 + 2^2 + 3^2 + ... + k^2 = \dfrac{k(k + 1)(2k + 1)}{6}...eqn.(1)$
Induction Step Assume $n = k + 1$ This is the $(k + 1)st$ term $ 1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = \dfrac{(k + 1)[(k + 1) + 1][2(k + 1) + 1]}{6} \\[5ex] 1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = \dfrac{(k + 1)(k + 1 + 1)(2k + 2 + 1)}{6} \\[5ex] 1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = \dfrac{(k + 1)(k + 2)(2k + 3)}{6}...eqn.(2) $
Induction Step	$ \color{black}{1^2 + 2^2 + 3^2 + ... + k^2} + (k + 1)^2 \\[3ex] \text{From } \color{black}{eqn. (1)}, \text{substitute} \\[3ex] \color{black}{\dfrac{k(k + 1)(2k + 1)}{6}} + (k + 1)(k + 1) \\[5ex] \dfrac{(k^2 + k)(2k + 1)}{6} + k^2 + k + k + 1 \\[5ex] \dfrac{2k^3 + k^2 + 2k^2 + k}{6} + k^2 + 2k + 1 \\[5ex] \dfrac{2k^3 + 3k^2 + k}{6} + \dfrac{k^2 + 2k + 1}{1} \\[5ex] \dfrac{1(2k^3 + 3k^2 + k) + 6(k^2 + 2k + 1)}{6} \\[5ex] \dfrac{2k^3 + 3k^2 + k + 6k^2 + 12k + 6}{6} \\[5ex] \dfrac{2k^3 + 9k^2 + 13k + 6}{6} \\[5ex] ............................................... \\[3ex] \underline{\text{Numerator}} \\[3ex] f(k) = 2k^3 + 9k^2 + 13k + 6 \\[3ex] \text{Try } k = -1 \\[3ex] f(-1) = 2(-1)^3 + 9(-1)^2 + 13(-1) + 6 \\[3ex] = 2(-1) + 9(1) - 13 + 6 \\[3ex] = -2 + 9 - 13 + 6 \\[3ex] = 0 \\[3ex] \implies \\[3ex] k = -1 \text{ is a zero} \\[3ex] (k + 1) \text{ is a factor} \\[3ex] \text{To find the remaining factors, let us use Synthetic Division} \\[3ex] \underline{\text{Synthetic Division}} \\[3ex] \begin{array}{r\|rrrr} -1 & 2 & 9 & 13 & 6 \\ + & & -2 & -7 & -6 \\ \hline & 2 & 7 & 6 & 0 \\ \end{array} \\[7ex] \text{Quotient} = 2k^2 + 7k + 6 \\[3ex] \implies \\[3ex] 2k^3 + 9k^2 + 13k + 6 = (k + 1)(k + 2)(2k + 3) \\[3ex] ............................................... \\[3ex] \dfrac{(k + 1)(k + 2)(2k + 3)}{6} $	$ \dfrac{(k + 1)(k + 2)(2k + 3)}{6} $
$LHS = RHS$ The formula is proved.

3rd Approach: Difference of Cubes leading to Telescoping Sum.
Assume the term: $(p + 1)^3$
Let us use the Pascal's Triangle to expand it. You may expand it using your preferred approach. $$ \begin{array}{c} 1 \\[2ex] 1 \quad 1 \\[2ex] 1 \quad 2 \quad 1 \\[2ex] 1 \quad 3 \quad 3 \quad 1 \\[2ex] \end{array} $$ $ (p + 1)^3 \\[3ex] = p^3 + 3p^2(1) + 3p(1)^2 + 1^3 \\[3ex] = p^3 + 3p^2 + 3p + 1 \\[5ex] (p + 1)^3 - p^3 = 3p^2 + 3p + 1 \\[3ex] $

$(p + 1)^3 - p^3 = 3p^2 + 3p + 1$
For	LHS	RHS
$p = 1$	$ (1 + 1)^3 - 1^3 \\[3ex] 2^3 - 1^3 $	$ 3(1)^2 + 3(1) + 1 $
$p = 2$	$ (2 + 1)^3 - 2^3 \\[3ex] 3^3 - 2^3 $	$ 3(2)^2 + 3(2) + 1 $
$p = 3$	$ (3 + 1)^3 - 3^3 \\[3ex] 4^3 - 3^3 $	$ 3(3)^2 + 3(3) + 1 $

$p = n$	$ (n + 1)^3 - n^3 $	$ 3(n)^2 + 3(n) + 1 $
Sum, S	$ (2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + ... + [(n + 1)^3 - n^3] \\[3ex] \text{This is now a telescoping sum.} \\[3ex] 2^3 - 1^3 + 3^3 - 2^3 + 4^3 - 3^3 + ... + (n + 1)^3 - n^3 \\[3ex] (n + 1)^3 - 1^3 \\[3ex] n^3 + 3n^2 + 3n + 1 - 1 \\[3ex] n^3 + 3n^2 + 3n $	$ [3(1)^2 + 3(1) + 1] \\[3ex] + [3(2)^2 + 3(2) + 1] \\[3ex] + [3(3)^2 + 3(3) + 1] \\[3ex] + ... + [3(n)^2 + 3(n) + 1] \\[5ex] 3[1^2 + 2^2 + 3^2 + ... + n^2] \\[3ex] + 3[1 + 2 + 3 + ... + n] \\[3ex] + 1 + 1 + 1 + ... + 1 \\[5ex] ................................. \\[3ex] 1^2 + 2^2 + 3^2 + ... + n^2 = S_{n^2} \\[3ex] 1 + 2 + 3 + ... + n = S_n \\[3ex] S_n = \dfrac{n(n + 1)}{2} ...\text{Sum of the first n natural numbers} \\[5ex] 1 + 1 + 1 + ... + 1 = n ...\text{The sum of 1 repeated n times is n} \\[3ex] ................................. \\[3ex] 3S_{n^2} + 3S_n + n \\[3ex] 3S_{n^2} + 3\left[\dfrac{n(n + 1)}{2}\right] + n \\[5ex] 3S_{n^2} + \dfrac{3n(n + 1)}{2} + n $
$ \text{LHS = RHS} \\[3ex] n^3 + 3n^2 + 3n = 3S_{n^2} + \dfrac{3n(n + 1)}{2} + n \\[5ex] 3S_{n^2} = n^3 + 3n^2 + 3n - \dfrac{3n(n + 1)}{2} - n \\[5ex] 3S_{n^2} = n^3 + 3n^2 + 2n - \dfrac{3n(n + 1)}{2} \\[5ex] LCD = 2 \\[3ex] 6S_{n^2} = 2n^3 + 6n^2 + 4n - 3n(n + 1) \\[3ex] 6S_{n^2} = 2n^3 + 6n^2 + 4n - 3n^2 - 3n \\[3ex] 6S_{n^2} = 2n^3 + 3n^2 + n \\[3ex] 6S_{n^2} = n(2n^2 + 3n + 1) \\[3ex] ........................................ \\[3ex] 2n^2 + 3n + 1 \\[3ex] 2n^2 + 2n + n + 1 \\[3ex] 2n(n + 1) + 1(n + 1) \\[3ex] (n + 1)(2n + 1) \\[3ex] ........................................ \\[3ex] 6S_{n^2} = n(n + 1)(2n + 1) \\[3ex] S_{n^2} = \dfrac{n(n + 1)(2n + 1)}{6} $

(2.) Find the smallest integer $n \gt 1$ so that $\sqrt{\dfrac{1^2 + 2^2 + 3^2 + ... + n^2}{n}}$ is an integer.

$ \dfrac{1^2 + 2^2 + 3^2 + ... + n^2}{n} \\[5ex] = (1^2 + 2^2 + 3^2 + ... + n^2) \div n \\[3ex] ................................................. \\[3ex] \text{Sum of the first n squares} \\[3ex] 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6} \\[5ex] ................................................. \\[3ex] = \dfrac{n(n + 1)(2n + 1)}{6} * \dfrac{1}{n} \\[5ex] = \dfrac{(n + 1)(2n + 1)}{6} \\[5ex] \implies \\[3ex] \sqrt{\dfrac{(n + 1)(2n + 1)}{6}} \\[5ex] \underline{\text{Testing: Trial and Error Approach}} \\[3ex] n \gt 1 \\[3ex] \text{Test } n = 2 \hspace{3em} \sqrt{\dfrac{(2 + 1)[2(2) + 1]}{6}} = \sqrt{\dfrac{3(5)}{6}} = \sqrt{\dfrac{15}{6}} \text{ is not an integer} \\[5ex] \text{Test } n = 3 \hspace{3em} \sqrt{\dfrac{(3 + 1)[2(3) + 1]}{6}} = \sqrt{\dfrac{4(7)}{6}} = \sqrt{\dfrac{28}{6}} \text{ is not an integer} \\[5ex] \text{Test } n = 4 \hspace{3em} \sqrt{\dfrac{(4 + 1)[2(4) + 1]}{6}} = \sqrt{\dfrac{5(9)}{6}} = \sqrt{\dfrac{45}{6}} \text{ is not an integer} \\[5ex] \text{Test } n = 5 \hspace{3em} \sqrt{\dfrac{(5 + 1)[2(5) + 1]}{6}} = \sqrt{\dfrac{6(11)}{6}} = \sqrt{11} \text{ is not an integer} \\[5ex] $ This is a long task... to keep testing until we find an integer.
We have to find a better approach.

$ \sqrt{\dfrac{(n + 1)(2n + 1)}{6}} = \text{some integer, } p \\[5ex] \dfrac{(n + 1)(2n + 1)}{6} = p^2 \\[5ex] (n + 1)(2n + 1) = 6p^2 \\[3ex] 2n^2 + n + 2n + 1 = 6p^2 \\[3ex] 2n^2 + 3n + 1 - 6p^2 = 0 \\[3ex] 2n^2 + 3n + (1 - 6p^2) = 0 \\[3ex] \text{Compare to the standard form: } an^2 + bn + c = 0 \\[3ex] a = 2 \\[3ex] b = 3 \\[3ex] c = 1 - 6p^2 \\[3ex] \underline{\text{Quadratic Formula}} \\[3ex] n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] \text{Because } n \gt 1 \text{ and } \sqrt{b^2 - 4ac} \text{ must be a positive integer}; \\[3ex] n = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \\[5ex] ................................................. \\[3ex] \text{Discriminant} = b^2 - 4ac \\[3ex] = 3^2 - 4(2)(1 - 6p^2) \\[3ex] = 9 - 8(1 - 6p^2) \\[3ex] = 9 - 8 + 48p^2 \\[3ex] = 1 + 48p^2 \\[3ex] ................................................. \\[3ex] $ The discriminant must be a perfect square.
This is because the square root of a perfect square is a positive integer.
In other words, the square root of the discriminant must be a positive integer.

$ \sqrt{1 + 48p^2} = \text{some integer, } k \\[3ex] ................................................. \\[3ex] \implies \\[3ex] n = \dfrac{-3 + \sqrt{1 + 48p^2}}{2(2)} \\[5ex] n = \dfrac{-3 + k}{4} \\[5ex] ................................................. \\[3ex] 1 + 48p^2 = k^2 \\[3ex] k^2 = 1 + 48p^2 \\[3ex] k^2 - 48p^2 = 1 \\[3ex] k^2 - 3 * 16 * p^2 = 1 \\[3ex] k^2 - 3 * 4^2 * p^2 = 1 \\[3ex] k^2 - 3(4p)^2 = 1 \\[3ex] $ This is now a Pell equation (Diophantine equation of the form: $x^2 - dy^2 = 1$ where d is a natural number and d is not a perfect square).
Solving it using this approach is the preferred approach if solving it manually (without using a Computational software).

$ \underline{\text{Pell Equation}} \\[3ex] x^2 - dy^2 = 1 \\[3ex] \text{A pair of positive integers, } (x, y) \text{ solves the Pell equation if} \\[3ex] \text{1st: } \exists d \in \mathbb{N}, \hspace{2em} d \text{ is not a perfect square} \\[3ex] \text{2nd: } \exists e \in \mathbb{N} \hspace{1em}:\hspace{1em} x + y\sqrt{d} = (a + b\sqrt{d})^e \\[3ex] \text{where } (a, b) \text{ is the fundamental solution: solution when }x, y \gt 0 \text{ and are minimal} \\[3ex] \text{So, for } k^2 - 3(4p)^2 = 1 \\[3ex] \text{Let us find the fundamental solution} \\[3ex] \underline{\text{Fundamental Solution}} \\[3ex] \text{When } p = 1 \\[3ex] k^2 = 1 + 3[4(1)]^2 \\[3ex] k = \sqrt{1 + 3(16)} \\[3ex] k = \sqrt{1 + 48} \\[3ex] k = \sqrt{49} \\[3ex] k = 7 \\[3ex] \text{Fundamental solution} = (k. p) = (7, 1) \\[5ex] \underline{\text{General Solutions}} \\[3ex] k + 4p\sqrt{d} = (7 + 4(1)\sqrt{3})^e \\[3ex] k + 4p\sqrt{d} = (7 + 4\sqrt{3})^e \\[3ex] d = 3 ...\text{is already an integer constant} \\[3ex] \text{We shall now focus on finding } n \text{ in cases where } k, p \text{ are integers} \\[3ex] n \text{ must also be an integer} \\[3ex] \text{The first instance of integer, } n \gt 1 \text { will be the answer} \\[3ex] \text{Recall that: } n = \dfrac{-3 + k}{4} \\[5ex] \text{When } e = 1 \\[3ex] (7 + 4\sqrt{3})^1 \\[3ex] = 7 + 4\sqrt{3} \\[3ex] k = 7 \\[3ex] 4p = 4 \\[3ex] p = \dfrac{4}{4} \\[5ex] p = 1 \\[3ex] n = \dfrac{-3 + 7}{4} \\[5ex] n = \dfrac{4}{4} \\[5ex] n = 1 \\[3ex] \text{But } 1 \ngtr 1 \\[5ex] \text{Next} \\[3ex] \text{When } e = 2 \\[3ex] (7 + 4\sqrt{3})^2 \\[3ex] = (7 + 4\sqrt{3})(7 + 4\sqrt{3}) \\[3ex] = 49 + 28\sqrt{3} + 28\sqrt{3} + 48 \\[3ex] = 97 + 56\sqrt{3} \\[3ex] k = 97 \\[3ex] 4p = 56 \\[3ex] p = \dfrac{56}{4} \\[5ex] p = 14 \\[3ex] n = \dfrac{-3 + 97}{4} \\[5ex] n = \dfrac{94}{4} ...\text{not an integer} \\[5ex] \text{Next} \\[3ex] \text{When } e = 3 \\[3ex] (7 + 4\sqrt{3})^3 \\[3ex] = (97 + 56\sqrt{3})(7 + 4\sqrt{3}) \\[3ex] = 679 + 388\sqrt{3} + 392\sqrt{3} + 672 \\[3ex] = 1351 + 780\sqrt{3} \\[3ex] k = 1351 \\[3ex] 4p = 780 \\[3ex] p = \dfrac{780}{4} \\[5ex] p = 195 \\[3ex] n = \dfrac{-3 + 1351}{4} \\[5ex] n = \dfrac{1348}{4} \\[5ex] n = 337 \\[3ex] 337 \gt 1 \\[3ex] \therefore n = 337 $

(10.) Suppose that X and Y are angles with $\tan X = \dfrac{1}{m}$ and $\tan Y = \dfrac{a}{n}$ for some positive integers a, m, and n.
Determine the number of positive integers $a \le 50$ for which there are exactly 6 pairs of positive integers (m, n) with $X + Y = 45^\circ$.

$ \underline{\text{Hint:}} \tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}...\text{Sum and Difference Formulas} \\[5ex] $

$ \text{Let us solve for }a \\[3ex] \tan (X + Y) = \dfrac{\tan X + \tan Y}{1 - \tan X \tan Y}...\text{Sum Formulas} \\[5ex] X + Y = 45^\circ \\[3ex] \tan X = \dfrac{1}{m} \\[5ex] \tan Y = \dfrac{a}{n} \\[5ex] \tan 45^\circ = \left(\dfrac{1}{m} + \dfrac{a}{n}\right) \div \left[1 - \left(\dfrac{1}{m}\right) * \left(\dfrac{a}{n}\right)\right] \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) \div \left(1 - \dfrac{a}{mn}\right) \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) \div \left(\dfrac{mn - a}{mn}\right) \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) * \left(\dfrac{mn}{mn - a}\right) \\[5ex] 1 = \dfrac{n + am}{mn - a} \\[5ex] n + am = mn - a \\[3ex] am + a = mn - n \\[3ex] a(m + 1) = n(m - 1) \\[3ex] a = \dfrac{n(m - 1)}{m + 1} ...eqn.(1) \\[5ex] \implies \\[3ex] \dfrac{n(m - 1)}{m + 1} \le 50 \\[5ex] $ Solving this inequality: two variables in one inequality is time-consuming...we do not know the value of either m or n
Even though we know that a, m, and n are positive integers, the guesses will take time.
So, let's see whether we can find a theorem that works in Number Theory to help us solve the question.

$ \text{From:} \\[3ex] n + am = mn - a \\[3ex] a = mn - am - n \\[3ex] a = m(n - a) - n \\[3ex] \text{To enable us to get a product of factors: add a to both sides} \\[3ex] a + a = m(n - a) - n + a \\[3ex] 2a = m(n - a) - 1(n - a) \\[3ex] 2a = (n - a)(m - 1) \\[3ex] $ To be continued.