Just 2 problems

[tranquil horizon]

Please solve these two problems they are not complicated for any of you trust me

[eager radish]

yes they arent complicated

why dont you do them yourself

they from the first imo

[limber ice]

You can prove that the GCD(21n+4, 14n+3) = 1

[pale vale]

For second question, you can do it violently. Just square both sides

It’s feasible and it seems to be an obvious solution

[pale vale]

For the first question

Set those disgusting stuff as k( k is a random natural number) and you’ll get an equation 21n+4 = k(14n+3)

After doing so, move the equation to get n = (3k-4)/7(3-2k)

So (3k-4)/(3-2k) must be a multiplier of 7

But however, no matter what integers you plug in, (3k-4)/(3-2k) remains negative

Therefore, our initial assumption was false

Done

@tranquil horizon

!done

[orchid geyser]

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