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Suppose that $n$ is a positive integer that satisfies the following system $(X)$:

$$n=60k_1+1 ;$$ $$n=7k_2 ;$$ for some integers $k_1, k_2$.

(Or you can write it in modular arithmetic form)

The question is: What is the general formula that generates all the positive integers that satisfy the two equations above?

My attempt: I have found the following formula that generates an infinite number of solutions:

$n=60(5+7t)+1$, where $t=0, 1, 2, 3, ...$

But how to prove that my formula produces all the possible solutions for the system $(X)$ above? Or, if not, then can anyone find a counter-example that solves $(X)$, but that can't be written in the form that I have found?

Bill Dubuque
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3 Answers3

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Note that your solution can also be written, by distributing the multiplication, as those numbers of the form $301 + 420t$ - that is, those $n$ that are congruent to $301$ mod $420$. Thus, it suffices to show that for every pair of solutions $n$ and $n'$ to your equations, their difference is a multiple of $420$.

To do this, suppose that you have two solutions: $$n=60k_1+1$$ $$n=7k_2$$ $$n'=60k_1'+1$$ $$n'=7k_2'$$ Then, their difference has the property that: $$n-n'=60(k_1-k_1')$$ $$n-n'=7(k_2-k_2').$$ In particular, their difference is a multiple of both $60$ and $7$. This is equivalent to saying that their difference is a multiple of the least common multiple of $60$ and $7$, which is $420$.

More generally, this argument suffices to show that any solution to a set of equations of the form $n\equiv a_i\pmod{m_i}$ is unique mod the LCM of the moduli - and if you're a bit careful, you can also use this as a lemma to show that if the moduli are pairwise coprime, a solution must also exist - which is the Chinese Remainder Theorem. (Though this is not how you would usually show existence because it sketches a non-constructive proof for something that can be solved constructively)

Milo Brandt
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$7|n=60k_1+1=7(8k_1)+4k_1+1\iff$

$7|4k_1+1\iff$

$ k_1\in \{7k+5: k\in \Bbb Z\}\iff $

$n=60k_1+1\in \{60(7k+5)+1: k\in \Bbb Z\}=\{420k+301: k\in \Bbb Z\}.$

DanielWainfleet
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    $7|4k_1+1\iff$ $ 7|2(4k_1+1)=7k_1+k_1+2 \iff $ $7|k_1+2\iff$ $ k_1+2\in {7(k+1):k\in \Bbb Z}\iff$ $ k_1\in {7(k+1)-2: k\in \Bbb Z}={7k+5: k\in \Bbb Z}.$ – DanielWainfleet Mar 29 '20 at 19:04
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The Chinese Remainder theorem says that if the moduli are coprime there will be a solution and the solutions will recur at the product of the individual moduli. Here that is $420$, which is the recurrence in your solution.

Ross Millikan
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  • Because there are $420$ choices for the remainders and each one has to have a solution, there must be only one solution $\bmod 420$ – Ross Millikan Mar 29 '20 at 16:10