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Where can I find the solution to BMO problems?

There is this problem bugging me where I have to find the smallest positive integer a such that

24a^2 = (b+1)(b-1)

for integers a and b, where a =/= 1
and b must be 1 mod4
and I'm not quite sure how to go about it

(edited 8 years ago)

Reply 1

catastrophe exam

haha brute forced it with a table of values... pretty sure that wasn't the way I was supposed to do it though

Reply 2

Renzhi10122

Original post by catastrophe exam

There is this problem bugging me where I have to find the smallest positive integer a such that

24a^2 = (b+1)(b-1)

for integers a and b, where a =/= 1
and b must be 1 mod4
and I'm not quite sure how to go about it

It's a Pell equation i.e.

b^2-24a^2=1

and one solution is

b=5, a=1

but we cannot have this.
So, we write the equation as

[br](5 - \sqrt{24} )(5+\sqrt{24} )=1[br](5 - \sqrt{24} )^2(5+\sqrt{24} )^2=1[br](49 - 10\sqrt{24} )(49+10\sqrt{24} )=1[br]49^2-24*10^2=1

so our solution is

a=10

I'll assume that you haven't met Pell equations before (most people haven't, and I haven't learnt them fully yet) but in essence, this technique gives us every solution, so a=10 is indeed the smallest.

I mean, I think a table of values is the simplest in this case, 10 is a small number.

Btw, which question is this, I don't recognise it.