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\(\large \color{black}{\begin{align} &\normalsize \text{Let a,b,c and d be four integers such that}\hspace{.33em}\\~\\ &a+b+c+d=4m+1\hspace{.33em}\\~\\ &k=a^2+b^2+c^2+d^2 \hspace{.33em}\\~\\ &\normalsize \text{where m is a positive integer}\hspace{.33em}\\~\\ &\normalsize \text{which one of the following is necessarily true}\hspace{.33em}\\~\\ & a.)\ \normalsize \text{The minimum possible value of k is }\ 4m^2-2m+1 \hspace{.33em}\\~\\ & b.)\ \normalsize \text{The minimum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ & c.)\ \normalsize \text{The maximum possible value of k is }\ 4m^2-2m+1 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{The maximum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ \end{align}}\)
minimum value of m is \(1\) so \(k_{min}=5\)
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lol i mean \((a+b+c+d)_{min}=5\)
not \(k_{min}\)
cuz \(m\) is a positive integer
yeah but m is given + integer
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.
@Zarkon , please
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)
i didnt understand , answer given is option (b.) , btw m not clever
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.
unlucky me, i didnt studied calculus yet
can a b c d repeat ?
Like a b c d can be literraly any integers?
yes \(\Large \{a,b,c,d\}\in \mathbb{Z}\)
\(k\) can grow infinitely, because you can choose them such that\[a=-b \\ c+d=4m+1\]and if you increase \(a\), \(k\) grows infinitely
So you just need to look for minimum value of \(k\)
yes i need to eliminate one option from 2
Now as you said, you just need to eliminate. You can do that by letting \(m = 1\).
\(a+b+c+d=5\), if \(m=1\)
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 \(a, b, c, d = m + \frac{1}{4}\).
\[k = (4m+1)^2 - 2 \cdot 6(m+ 1/4)^2\]\[= (4m + 1)^2 - (12m^2 + 6m + 3/4)\]\[= 4m^2 + 2m + 1/4\]So the minimum of \(k\) for all real numbers is \(4m^2 + 2m + 1/4\).
So it of course cannot hold the value \(4m^2 - 2m + 1\), thus the answer is \(4m^2 + 2m + 1\).
Is there some way to use AM-GM-HM here?
but how can it be \(m+\dfrac14\) as all variables are given as integers
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).
u maximized \(k\) and this is \(k_{max}\) right ? \(k_{max} = (4m+1)^2 - 2 \cdot 6(m+ 1/4)^2\)
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 \(m+1\).
Yes! This is \(k_{max}\) for real numbers.
but as mukushla pointed out the \(k_max\) can be up to infinity , i just need to work on 2 options for \(k_{min}\)
Sorry, I meant \(k_{min}\). That is the \(k_{min}\) for reals and it coincides with \(\lambda_{max}\).
Nice! with the options given, one way to get a quick answer is by plugging in \(m=1\) and see which option makes sense
for m=1 option a-> 3 and option b->7 ?
option a doesn't makes sense , so option b is correct . nice
I literally said the same thing in the first reply... :|
lol yea ,sry
thnks all.

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