Inductive Proof (+linear equation in four variables)

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SUMMARY

This discussion focuses on proving by induction that for all integers n ≥ 5, there exist non-negative integers m1 and m2 such that n = 2m1 + 3m2. The user has completed the base case and is working on the inductive step, which requires demonstrating that if n can be expressed in the form of 2m1 + 3m2, then n + 1 can also be expressed in the same form. The challenge lies in finding new integers m1' and m2' that satisfy the equation 2m1 + 3m2 + 1 = 2m1' + 3m2'. The user seeks a method to achieve this increment using only the coefficients 2 and 3.

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  • Understanding of mathematical induction
  • Familiarity with linear equations in multiple variables
  • Basic knowledge of number theory, specifically integer solutions
  • Experience using computational tools like Wolfram Alpha
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Mathematicians, students studying number theory, and anyone interested in proofs involving linear combinations and mathematical induction.

sweetreason
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I'm trying to prove by induction that [tex]\forall n \geq 5, \exists m_1, m_2 \in \mathbb{N}[/tex] such that [tex]n = 2m_1 +3m_2.[/tex]I've done the base case, and the inductive step boils down to showing that [tex]\exists m_1 \prime m_2 \prime[/tex] such that [tex]2m_1 +3m_2 +1 = 2m_1 \prime +3m_2 \prime[/tex]. Maybe I'm forgetting something from grade school algebra, but I have no idea how to solve for [tex]m_1 \prime, m_2 \prime[/tex]. I've plugged it into wolfram alpha [http://www.wolframalpha.com/input/?i=2*x_1+3*x_2+++1+=+2*y_1+++3*y_2] and got solutions (all I care about is the case n =1 for the integer solutions wolfram gives) but I want to know how to arrive there.
 
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If you have to add one to a number, how could you do that if you only have increments / decrements of 2 and 3 available?
 

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