how can i solve this and this type of equations

[winged fossil]

3*2^m + 1 =n^2. Find all (n:m)

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[prime nymph]

@winged fossil subtract 1 from both sides and you get $3*2^m=(n+1)(n-1)$. Now see how you can factor the left side of your equation.

[snow bolt]

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[winged fossil]

How?

Can i need to use (mod)

[prime nymph]

I don't think that will help

Either n+1 or n-1 has to be divisible by 3 so you have two cases:
n+1=3*2^a
n-1=2^b
a+b=m
or
n+1=2^a
n-1=3*2^b
a+b=m.

In the first case subtracting the two equations gives you 3*2^a-2^b = 2 and obviously a and b cant be greater or equal to 2 at the same time (because then 4|2) so that will give you a few solutions

The other case works pretty much the same

Where did you get this question from? @winged fossil

[winged fossil]

I am study olympiads and i get some teachers

I am from azerbaijan

[prime nymph]

That's very nice

[winged fossil]

My hope is succes in olympiads

[prime nymph]

I have also been trying that. It takes a lot of practise