Math Olympiad problem | How to solve for "m" & "n" in this problem? @MathOlympiad0

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Can You Solve For 'n'? | Math Olympiad Questions | A Beautiful Exponential Problem | Algebra problem
Can You Solve For 'n'? | Math Olympiad Questions | A Beautiful Exponential Problem | Algebra problem
Can You Solve For 'n'? | Math Olympiad Questions | A Beautiful Exponential Problem | Algebra problem
Can You Solve For 'n'? | Math Olympiad Questions | A Beautiful Exponential Problem | Algebra problem
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[@sanjeevshukla6865]

For this particular problem, why not just try combinations of 1 < m, n < 4 since 4^3 is 64?

[@mohinkhan2503]

5*3 and 3*7. 15+21=35 ?

[@marcgriselhubert3915]

As m and n are in Z+ and different from 1 they are superior or equal to 2. As 35< 4^3, m and n are strictly inferior to 4.
Only possible solutions: m and n are equal to 2 or 3, we check the four possibilities and find easily that (m,n) = (2,3) or (3,2), that' all.

[@walterwen2975]

Math Olympiad: m³ + n³ = 35, m, n ꞓZ⁺, m, n ≠ 1; m, n = ?
m, n positive integers, 6 > m + n > 1; m > n > 0 or n > m > 0
First method:
m³ + n³ = 35 = 3³ + 2³ = 2³ + 3³; m = 3, n = 2 or m = 2, n = 3
Second method:
m³ + n³ = (m + n)(m² - mn + n²) = 35 = (5)(7); 5, 7 are prime numbers
m² - mn + n² > m + n; m, n ꞓZ⁺, 6 > m + n > 1
m + n = 5, m² - mn + n² = 7; (m + n)² - (m² - mn + n²) = 3mn = 25 - 7 = 18
mn = 6; n = 6/m, m + n = m + 6/m = 5, m² - 5m + 6 = (m - 2)(m - 3) = 0
m - 2 = 0, m = 2, n = 6/m = 3 or m - 3 = 0, m = 3, n = 6/m = 2
Answer check:
m³ + n³ = 35; Confirmed in First method
Final answer:
m = 2, n = 3 or m = 3, n = 2