$h\mid (3a + 5b)$, prove $h\mid a$ and $h\mid b$

Question

I have this homework question.

"For any integer $a$ and $b$, prove that $\gcd(a,b) = \gcd(3a+5b,11a+18b)$."

I know that if $ g = \gcd(a,b)$

and

$h = \gcd(3a+5b,11a+18b)$

then

$g = h$

iff $g \leq h $ and $h \leq g$.

I successfully proved that $g \leq h$.

Now, to prove that $h \leq g$, I need to prove that $h\mid (a,b)$, but I can't seem to find how I should prove this.

$\because h = gcd(3a+5b,11a+18b) \Rightarrow h \mid (3a+5b)$

From here I'm stuck on how to get $a$ and $b$ seperate.

Any hint would be very helpful.

Edit: Since this question was marked duplicate and I was given these 1,2,3 links to check, I did check them and didn't find my answer because all of these questions have given that $gcd = 1$, whereas my question doesn't tell if $gcd = 1$ and furthermore these questions are a bit complex for me to understand since I'm a new learner of number theory.

I edited my question after checking other answers and they didn't give me my answer. — Ramsha Khalid, Oct 08 '18 at 19:55
You question is a special case of the linked dupesince $,5\cdot 11-3\cdot 18 = 1$. Most of the ideas in the other linked answers will also work here even though they deal only with the coprime case. Indeed, we can reduce to the coprime case by cancelling the gcd from both sides. — Bill Dubuque, Oct 08 '18 at 20:12

score 1 · Accepted Answer · answered Oct 08 '18 at 19:28

1

Since $$ h\mid 3a+5b\;\;\;{\rm and} \;\;\;h\mid 11a+18b$$ we have $$h\mid 4(3a+5b)-(11a+18b)= a+2b$$

Now $$h\mid 3(a+2b)-(3a+5b)=b$$

so $$h\mid (a+b)-2b = a$$

so $h\mid \gcd(a,b)$.

answered Oct 08 '18 at 19:28

nonuser

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$h\mid (3a + 5b)$, prove $h\mid a$ and $h\mid b$

1 Answers1