## mathmath333 one year ago The question

1. mathmath333

\large \color{black}{\begin{align} &\normalsize \text{Let a,b,c and d be four integers such that}\hspace{.33em}\\~\\ &a+b+c+d=4m+1\hspace{.33em}\\~\\ &k=a^2+b^2+c^2+d^2 \hspace{.33em}\\~\\ &\normalsize \text{where m is a positive integer}\hspace{.33em}\\~\\ &\normalsize \text{which one of the following is necessarily true}\hspace{.33em}\\~\\ & a.)\ \normalsize \text{The minimum possible value of k is }\ 4m^2-2m+1 \hspace{.33em}\\~\\ & b.)\ \normalsize \text{The minimum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ & c.)\ \normalsize \text{The maximum possible value of k is }\ 4m^2-2m+1 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{The maximum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ \end{align}}

2. mathmath333

minimum value of m is $$1$$ so $$k_{min}=5$$

3. Loser66

How can you get it?

4. mathmath333

lol i mean $$(a+b+c+d)_{min}=5$$

5. mathmath333

not $$k_{min}$$

6. mathmath333

cuz $$m$$ is a positive integer

7. mathmath333

yeah but m is given + integer

8. mathmath333

i was thinking to use this AM-GM inequality \large \color{black}{\begin{align} \dfrac{a^{2}+b^{2}+c^{2}+d^{2}}{4}\geq (a^{2}b^{2}c^{2}d^{2})^{(1/4)} \hspace{.33em}\\~\\ \end{align}} but $$a^{2}b^{2}c^{2}d^{2}$$ is not given.

9. Loser66

10. Zarkon

you can use mathematical induction if you are a little clever. (and you know which is the correct answer, which there is a way, with calculus, to eliminate 3 out of the 4 possible answers)

11. mathmath333

i didnt understand , answer given is option (b.) , btw m not clever

12. Zarkon

you can use Lagrange Multipliers to deduce that the only possible answer is b... then you can use induction on $$m$$ to show that $$4m^2+2m+1$$ is indeed the smallest $$k$$ can be.

13. mathmath333

unlucky me, i didnt studied calculus yet

14. SolomonZelman

can a b c d repeat ?

15. SolomonZelman

Like a b c d can be literraly any integers?

16. mathmath333

yes $$\Large \{a,b,c,d\}\in \mathbb{Z}$$

17. anonymous

$$k$$ can grow infinitely, because you can choose them such that$a=-b \\ c+d=4m+1$and if you increase $$a$$, $$k$$ grows infinitely

18. anonymous

So you just need to look for minimum value of $$k$$

19. mathmath333

yes i need to eliminate one option from 2

20. ParthKohli

@ganeshie8

21. ParthKohli

Now as you said, you just need to eliminate. You can do that by letting $$m = 1$$.

22. mathmath333

$$a+b+c+d=5$$, if $$m=1$$

23. ParthKohli

Let's try to find the minimum for real numbers a, b, c, d. I think we should use the expansion of $$(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + ac + ad + bd )$$. So$(4m+1)^2 = k + 2\lambda$$k = (4m + 1)^2 - 2\lambda$To maximise $$k$$, we minimise $$\lambda$$. Since we know that $$a + b + c + d = 4m + 1$$, we can see that $$\lambda$$ will be minimum for $$a, b, c, d = m + \frac{1}{4}$$.

24. ParthKohli

$k = (4m+1)^2 - 2 \cdot 6(m+ 1/4)^2$$= (4m + 1)^2 - (12m^2 + 6m + 3/4)$$= 4m^2 + 2m + 1/4$So the minimum of $$k$$ for all real numbers is $$4m^2 + 2m + 1/4$$.

25. ParthKohli

So it of course cannot hold the value $$4m^2 - 2m + 1$$, thus the answer is $$4m^2 + 2m + 1$$.

26. ParthKohli

Is there some way to use AM-GM-HM here?

27. mathmath333

but how can it be $$m+\dfrac14$$ as all variables are given as integers

28. ParthKohli

I started by solving this problem for real numbers as there is no clear way to do so for integers (maybe there is - @mukushla might know about it).

29. mathmath333

u maximized $$k$$ and this is $$k_{max}$$ right ? $$k_{max} = (4m+1)^2 - 2 \cdot 6(m+ 1/4)^2$$

30. ParthKohli

At our level, I don't think there's a satisfactory solution. As a rule, maximum/minimum occurs when all numbers are close to each other (possibly even equation). Here, I believe it occurs when three variables are $$m$$ and one of them is $$m+1$$.

31. ParthKohli

Yes! This is $$k_{max}$$ for real numbers.

32. mathmath333

but as mukushla pointed out the $$k_max$$ can be up to infinity , i just need to work on 2 options for $$k_{min}$$

33. ParthKohli

Sorry, I meant $$k_{min}$$. That is the $$k_{min}$$ for reals and it coincides with $$\lambda_{max}$$.

34. ganeshie8

Nice! with the options given, one way to get a quick answer is by plugging in $$m=1$$ and see which option makes sense

35. mathmath333

for m=1 option a-> 3 and option b->7 ?

36. mathmath333

option a doesn't makes sense , so option b is correct . nice

37. ParthKohli

I literally said the same thing in the first reply... :|

38. mathmath333

lol yea ,sry

39. mathmath333

thnks all.