Homework Help: Basic math Olympiad problem

1. May 6, 2017

vishnu 73

<Moderator's note: moved from a technical forum, so homework template missing.>

so this is the question.
i want to know if there is a solution without using calculus maybe trig substitution maybe other methods?
i tried trig substitition
i let u = √2 cosx
and v be sinx
am i on the right track

Last edited by a moderator: May 8, 2017
2. May 6, 2017

Staff: Mentor

You need two variables - with only x you cannot cover the whole plane you have to search.

3. May 8, 2017

vishnu 73

ok fine let
v be siny
but this only makes the problem complicated how am i supposed to minimum value of function with different variables

4. May 8, 2017

Staff: Mentor

It has to be a minimum with respect to both variables. While that is technically not sufficient to have a global minimum, it will do the job here.

5. May 8, 2017

vishnu 73

so terms containing x should be minimum themselves and terms containing y must be local minimum too?

6. May 8, 2017

Staff: Mentor

Yes, or the same with u and v without substitution. If a point is not a minimum with respect to these variables, there is a point nearby which has a lower value.

7. May 8, 2017

Staff: Mentor

Are you allowed to use a calculator??

8. May 9, 2017

no

9. May 9, 2017

vishnu 73

@mfb
i am just realizing i got back to using calculus and partial derivative i know that method already is there any other

10. May 9, 2017

Staff: Mentor

Sometimes there are other options. You have the sum of two squares. Both squares are not negative, so if there is a point where both squares are zero, or at least at a global minimum within the allowed parameter range, you found a global minimum. Then you just have to check if it is within the given range of the coordinates.
There is no such point in this problem (unless you allow imaginary arguments).

11. May 10, 2017

vishnu 73

so is the calculus the only method? without imaginary numbers only real because i have it done the calculus way already

12. May 10, 2017

Buffu

Did you try AM-GM ?

13. May 12, 2017

haruspex

One thing that makes it hard is having both variables in both terms. If you were to make a substitution to two new variables, x and y, so that one term only contains x then you can concentrate first on minimising the other term wrt y.
The downside is that checking the given bounds gets awkward.

Another possible start is to consider what happens on the boundaries.

Last edited: May 12, 2017
14. May 13, 2017

vishnu 73

@Buffu
how to apply am-gm here

@haruspex
yup i am finally seeing that calculus is the most straight forward approach here

15. May 13, 2017

haruspex

Standard differential calculus approach will lead to quartics to solve.
I think I have the answer, using my first suggestion. Write x+u for v and find the min wrt u. That should give you an expression for v in terms of u. It is rather nasty, but see if you can spot a fairly simple pair of values satisfying it.
Showing that produces the minimum is easier than finding it from calculus.

16. May 13, 2017

vishnu 73

17. May 13, 2017

Buffu

$\displaystyle (u - v)^2 + \left(\sqrt{2- u^2} - {9\over v}\right)^2 \ge 2(u-v)\left(\sqrt{2- u^2} - {9\over v}\right)$

Maybe do $v = ku$ now.

I don't think it simplifies the equation much but better than nothing, right ?

18. May 13, 2017

vishnu 73

okay i will give that a try too give me some time i am now doing another problem

19. May 13, 2017

haruspex

Very neat, and yes, that is the answer I got.
The method at that link does produce quartics - see the expression with both y and 1/y3 - but it turns out to factorise nicely.

20. May 13, 2017

vishnu 73

oh if thats what you meant by quartics then yes but anyways i will give your method a try too

21. May 21, 2017

vishnu 73

@haruspex

is this correct is this the method

would the method if i let u = x + v?

@Buffu
i am not really too sure how to prceed with am gm i get stuck once after i make the substituion please help me thanks

22. May 21, 2017

haruspex

That will do. Now minimise the function wrt v, i.e. x constant. Can you see how to simplify it before differentiating?

23. May 21, 2017

vishnu 73

thanks i am getting the exact same equation as before it works

but why does this method work?

24. May 21, 2017

haruspex

It is really the same as just differentiating the original equation partially wrt u and v, but by a change of axes it makes the algebra a bit easier.

25. May 27, 2017

vishnu 73

hi i have just one minor question here can you help me verify thanks ?

so this is the question
find all integer solutions to n4 +2n3 + 2n2 + 2n + 1 = m2
i factored it to become
(n+1)2(n2 +1) = m2
aren't the solution just (n,m) = (0,1) (-1 , 0)
is this correct?