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Tutor profile: Apratim D.

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Apratim D.
PhD candidate in Mathematics | Mathematics and CS/programming tutor for the past 8 years
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Questions

Subject: Linear Algebra

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Question:

Suppose $$A$$ is an $$n \times n$$ matrix with rational entries such that its minimal polynomial is $(\mu (x) = x^3+5x+2. $) Show that $$3$$ divides $$n$$.

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Apratim D.
Answer:

Note that $$\mu$$ is a polynomial with integer coefficients. The rational root theorem says that if $$\mu$$ has a rational root which can be written as $$p/q$$ in its lowest terms, then $$p$$ divides the constant term which is $$2$$ in this case and $$q$$ divides the leading coefficient which is $$1$$ in this case. That is, all rational roots of $$\mu$$ must be integral and moreover, they should be divisors of $$2$$. So, the candidates for the rational roots of $$\mu$$ are $$\{1,-1,2,-2 \}$$ but if we plug these back in the polynomial, we see that none of these are roots of $$\mu$$. Therefore, $$\mu$$ is irreducible over the rationals. Denote $$\chi(x) = \det (xI-A)$$ as the characteristic polynomial of $$A$$. $$\chi$$ has degree $$n$$. We know from a theorem in Linear Algebra that the minimal polynomial and the characteristic polynomial have the same irreducible factors.That is, if $$\mu = f_1^{a_1}\cdots f_k^{a_k}$$, then $$\chi = f_1^{b_1}\cdots f_k^{b_k}$$ where $$f_i$$ are irreducible polynomials and $$a_i \le b_i$$. Now, here since $$\mu$$ is irreducible that means it has only one irreducible factor - itself. Therefore we must have $(\chi(x) = (\mu (x))^r $) Therefore, $$n = \deg (\chi)= r\cdot\deg (\mu) = 3r$$.That is, $$3$$ divides $$n$$.

Subject: Number Theory

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Question:

Find all prime numbers $$p$$ greater than $$65$$ such that $$p-64$$ is the fourth power of an integer.

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Apratim D.
Answer:

So, we need $$p-64 = q^4 $$ for some integer $$q$$. Therefore, we have \begin{align*} p &= q^4+64\\ &= (q^2)^2 + 8^2\\ &=(q^2)^2 + 8^2 + 2 q^2\cdot8 - 2 q^2\cdot 8\\ &=(q^2+8)^2 - 16q^2\\ &= (q^2+8+4q)(q^2+8-4q) \end{align*} Now, $$q^2+8-4q = (q-2\sqrt{2})^2+ 2q\cdot2\sqrt{2} -4q = (q-2\sqrt{2})^2 + 4q(\sqrt{2}-1) > 4q(\sqrt{2}-1) > 4q\times 0.414 > 1.6q > 1$$. That is, $$q^2+8-4q$$ is an integer greater than $$1$$. That means we have written $$p = (q^2+8+4q)(q^2+8-4q) $$ as a product of two integers greater than $$1$$ which contradicts the fact that $$p$$ is a prime. Therefore, no such prime $$p > 65$$ exists such that $$p-64$$ is the fourth power of an integer.

Subject: Calculus

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Question:

Let us denote by $$f(x)$$, the following polynomial: $(f(x) = (x-1)^2\prod_{i=2}^{98}(x-i) $) What are the real roots of the polynomial $(p(x) = (f'(x))^2 - f''(x)f(x) \ ?$)

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Apratim D.
Answer:

For $$x \notin \{1,2, \ldots, 98 \}$$, we have $(\frac{f'(x)}{f(x)} = \frac{2}{(x-1)} + \sum_{i=2}^{98}\frac{1}{(x-i)} $) Differentiating the above expression, we get $(\frac{f''(x)f(x) - (f'(x))^2}{(f(x))^2} = \frac{-2}{(x-1)^2} - \sum_{i=2}^{98}\frac{1}{(x-i)^2}$) That is, $((f'(x))^2 - f''(x)f(x) = (f(x))^2\left[\frac {2}{(x-1)^2}+\sum_{i=2}^{98}\frac{1}{(x-i)^2} \right] $) For all real $$x \notin \{1,2, \ldots, 98 \}$$, the above expression is positive. So the possible real candidates for the above expression to be $$0$$ lie in $$\{1,2, \ldots, 98 \}$$. Let $$\alpha \in \{1,2, \ldots, 98 \}$$ be one such candidate. Then $$f(\alpha) = 0$$. Moreover, $((f'(\alpha))^2 - f''(\alpha)f(\alpha) = 0 $) That is, $( (f'(\alpha))^2 = 0$) So, we have $$f(\alpha) = f'(\alpha) = 0$$ which means such an $$\alpha$$ must be a repeated real root of $$f$$. $$f$$ has only one repeated real root which is $$1$$. Therefore, the only real root of $$p$$ is $$1$$.

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