The question

- mathmath333

The question

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- mathmath333

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

- mathmath333

minimum value of m is \(1\) so \(k_{min}=5\)

- Loser66

How can you get it?

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## More answers

- mathmath333

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

- mathmath333

not \(k_{min}\)

- mathmath333

cuz \(m\) is a positive integer

- mathmath333

yeah but m is given + integer

- 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.

- Loser66

@Zarkon , please

- 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)

- mathmath333

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

- 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.

- mathmath333

unlucky me, i didnt studied calculus yet

- SolomonZelman

can a b c d repeat ?

- SolomonZelman

Like a b c d can be literraly any integers?

- mathmath333

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

- 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

- anonymous

So you just need to look for minimum value of \(k\)

- mathmath333

yes i need to eliminate one option from 2

- ParthKohli

@ganeshie8

- ParthKohli

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

- mathmath333

\(a+b+c+d=5\), if \(m=1\)

- 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 \(a, b, c, d = m + \frac{1}{4}\).

- 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 \(4m^2 + 2m + 1/4\).

- ParthKohli

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

- ParthKohli

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

- mathmath333

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

- 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).

- mathmath333

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

- 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 \(m+1\).

- ParthKohli

Yes! This is \(k_{max}\) for real numbers.

- mathmath333

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

- ParthKohli

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

- 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

- mathmath333

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

- mathmath333

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

- ParthKohli

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

- mathmath333

lol yea ,sry

- mathmath333

thnks all.

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