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Question: Prove by Mathematical Induction, or disprove, that there must be a non-negative integral solution $x_1, x_2$ for the equation $3x_1 + 5x_2 = p$, for any natural number $p \geq 10$.

I am not sure how to handle $x_1$ and $x_2$, can they share the same value to prove like $3(k+2) + 5(k+2) \geq 10$?

Robert Shore
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cody
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  • (1) Please use MathJax whenever you type a formula on this site. It’s pretty easy. You just enclose your formulae into the $ signs, for the most part. (2) Welcome to MathSE! Please see how to ask a good question. It’s better to show your attempt in the post. What progress did you achieve? Where were you “stuck”? – Aig Apr 29 '24 at 08:32
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    Hint: If you decrease $x_2$ by $1$ and increase $x_1$ by $2$ then $3x_1+5x_2$ increases by $1$. – Kavi Rama Murthy Apr 29 '24 at 08:32
  • Does this answer your question? Frobenius coin problem. Or https://math.stackexchange.com/questions/8241/generating-integers-from-a-linear-combination-of-integers. Or https://math.stackexchange.com/questions/723356/frobenius-coin-problem-induction-help. Or https://math.stackexchange.com/questions/4291632/the-frobenius-coin-problem-constructive-proof. Or https://math.stackexchange.com/questions/1181222/proof-that-every-number-%e2%89%a5-8-can-be-represented-by-a-sum-of-fives-and-threes. – Arthur Apr 29 '24 at 08:58
  • Or https://math.stackexchange.com/questions/3831585/frobenius-coin-problem-using-mathematical-induction. Or https://math.stackexchange.com/questions/1500981/chicken-mcnugget-theorem-frobenius-coin-problem. Or https://math.stackexchange.com/questions/563401/frobenius-coin-problem-5-and-9. – Arthur Apr 29 '24 at 09:02

1 Answers1

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Let us say, this is true for some $p$.

$$3x_1+5x_2 = p$$

We need to show, that this is true for $p+1$.

Hint: Notice that $3.2-5.1 = 1$ and $5.2-3.3 = 1$

Handwaving Solution:

$$3x_1+5x_2+1 = p+1$$ Replace 1 with $3.2-5.1$

$$3(x_1+2)+5(x_2-1) = p+1$$

But this decreases $x_2$, so it may happen that $x_2$ becomes negative.

Then, $3(x_1-3)+5(x_2+2) = p+1$ works.

Notice that it can not happen that $x_1$ also becomes negative,because $x_2$ becoming negative in the first case implies that $x_2$ is zero so, as p is atleast 10, $x_1$ has to be greater than 3.So, $x_1-3$ is positive always in the second case.

I have left the hypothesis and middle work to construct a rigourous proof upto you.

Krave37
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