induction: for all integers n>=1 , 2(*7^n)- (3*5^n) + 1 is divisible by 24

[fervent vine]

What did you try

[terse lichen]

I tried adding zero like a multiple of 5^k

[fervent vine]

$$24 \mid 2\cdot 7^n - 3\cdot 5^n + 1$$

[hollow bay]

casework

[terse lichen]

Yeah I got that subbed in

[fervent vine]

This is the Q just to be sure?

[terse lichen]

Yes

[fervent vine]

I mean its induction

Base holds

Make the assumption

Basically the step now

$$24 \mid 2 \cdot 7^{n + 1} - 3\cdot 5^{n + 1} + 1$$

[hollow bay]

casework

[fervent vine]

How much induction do you need to use?

Like can we simplify this in to something simple with the assumption and use modular arithmetic

[terse lichen]

That’s what I tried

[fervent vine]

Well you could easily prove
$24 \mid 6(5^n - 1)$

[hollow bay]

casework

[terse lichen]

That part I was confused on ^

Proving that

[fervent vine]

I mean its pretty trivial

$4 \mid 5^n - 1$

[hollow bay]

casework

[fervent vine]

As $5^n \equiv 1^n \equiv 1 \pmod 4$

[hollow bay]

casework

[fervent vine]

But im guessing you cant use that

[terse lichen]

Probably not I’ve never seen that

[fervent vine]

Ok i think i got it

$24 \mid 14 \cdot 7^n - 15 \cdot 5^n + 1$
Notice how this is
$24 \mid 2\cdot 7^n - 3\cdot 5^n + 1 + 12(7^n - 5^n)$

[hollow bay]

casework

[fervent vine]

Only thing you have to do now is prove that $7^n - 5^n$ is even

[hollow bay]

casework

[fervent vine]

And that doesnt really need proving

Its odd - odd

[terse lichen]

I can do that

I’ll give it a try thanks

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[spice plover]

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[terse lichen]

!solved

[candid basin]

.solved