At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Because 1 is a little bigger than 0?...

Confused!!!

ah nvm scratch this stuff i dont think its the right way

dan I think the math is just off. Use used f(2) = 2 :)

oh lets use this we can use that induction step after all

edit: I meant 2009 = 49*41 of course.

oohhh haha ok that makes a little bit more sense lol

|dw:1441425591667:dw|

|dw:1441425872198:dw|

u can get a base case and solve it

i think u can get a closed form from that too now

your equations aren't right :(
|dw:1441426034574:dw|

wait how comes thats wrong
f(x-y)=sqrt(f(x*y)+1) right

|dw:1441426260927:dw|

Oh, sorry I thought the LHS was n*1. Better reread.

ah gotcha

can we do some telescoping stuff here though

|dw:1441426410790:dw|

we dont have f(1) or anythign so what can we do

Still, you have 4 unknowns with 3 equations. You want k equations with k unknowns for the base case.

Anyway, I have a solution strategy, but it's not pretty.

oh cool that works

f(2) = [f(1)]^2 - 1, that is.
BTW, from the last one at least we get f(5) = f(1)

There must be a simpler way than just brute forcing your way through.

ill work on finding closed form for this it looks kind of fun
|dw:1441427086013:dw|

oh serious

assuming p1>q1, p2>q2 of course.
BTW this is looking more and more like a number theory problem.

yea

like we know how many expreresion can exist for each f(k)

if k=xy i mean not the x-y part

I think f(x) is just the constant function.

Close the thread :)

I was kind of playing with this more I think f is a constant function.

thatn is simply brilliant !

x - y = u
xy = u
does that allow us to get the closed form too

so solving that quadratic function will give us a closed form

oh it does, nice..

well I might have a question...

\[\sqrt{c^2+4c}>c \text{ is this true for all positive real } c \\\]

definitely

yeah you are right

I just wanted to verify the choice of y would always be positive for positive real c

\(\sqrt{c^2} = c\), and \(\sqrt(x)\) is increasing, so yeah.

oh yeah duh
thanks @beginnersmind