5

If $x$ and $y$ are identical, then $x^2+x=x(x+1)$ has to be a square where $x^2<x(x+1)<(x+1)^2$. If $x$ and $y$ are different, then wlog $y<x$ and then $x^2<x^2+y<x(x+1)<(x+1)^2$. There may be a problem here ... .


4

If $Ax=u$ has no solution, it is also true that $Ax = -u$ has no solution. So we are saying your $v=-u$ Next, $Ax = u + v = u +(-u) = u-u=0$ always has a solution


3

The first equation is $xb+ya = ab$, while the second one is $a^3y - b^3x = xy(b^2-a^2).$ Let's rearrange the second one as $ay(a^2+ax) = bx(b^2+by).$ Using the first one, we get $a(a+x)(ab-xb) = b(b+y)(ab-ya)$. Thus, $(a+x)(a-x) = (b+y)(b-y)$, which implies the thesis.


3

Rewrite the two given equations as $$ ab-2cd=3 \quad\text{and}\quad ac+bd=1, $$ and note for later use that $a\ne0$ and $b\ne0$ and that at most one $c$ and $d$ can equal $0$, and also that at least one variable is even. Squaring both sides of both equations and rearranging yields $$ a^2b^2+4c^2d^2=9+4abcd \quad\text{and}\quad a^2c^2+b^2d^2=1-abcd, $$ and ...


2

If $x^2+y=m^2\Rightarrow m\geq x+1\Rightarrow y\geq 2x+1$ . This shows that if $y^2+x$ is a square then $x\geq 2y+1$. Combine the last two inequalities to see that $y\geq2x+1\geq 2(2y+1)+1=4y+3$ which is of course impossible.


1

Luca's solution works, and is likely how they intended this to be solved. I'd like to point out that OP's approach of "substitute the first equation into the second" can be forced through to completion. We have $y = \frac{b(a-x) } { x } $ from the first equation. Substituting into the second equation, we get $ \frac{a^3}{x} - \frac{ab^2}{a-x} = b^...


1

You consider the function $\|x(t)\|_2$. If you want to show that it is increasing, then it's enough to show that $\|x(t)\|_2^2$ is increasing, since $\|x(t)\|_2$ is non-negative and the square-root with this co-domain is increasing. However, $\|x(t)\|_2^2 = x(t)^T x(t)$. This is a product of differentiable functions of $t$, hence differentiable. We must use ...


1

Welcome to MSE! I think some of the confusion regarding the responses to this question come from the (admittedly subtle) distinction between a closed form solution to a recurrence and an extension of a recurrence (to all of $\mathbb{C}$, say). If you're able to solve a recurrence, so that $T(n) = f(n)$ where $f$ is some "nice" function, then we get ...


1

You claim that The only continuous function F that satisfies the above system of equations is, as we probably all know, $$ F(n) = \frac{\varphi^n - \psi^{n}}{\sqrt{5}}, $$ This claim in false even if $\,F\,$ is restricted to real functions. Given any integer $\,k,\,$ define the confinuous complex function $$ F_k(x) := e^{2\pi i k x}\frac{\varphi^x - \psi^{...


1

Let me use an example with $t=6$. $$\frac {x^5}{\prod_{n=1}^5 \left(1-\frac{x}{a_n}\right)}=\sum_{n=1}^5 \frac {c_n}{1-\frac{x}{a_n}}$$ Cross multiply to have $$x^5=\Bigg[\sum_{n=1}^5 \frac{c_n}{1-\frac{x}{a_n}}\Bigg]\,\times\, \Bigg[\prod_{n=1}^5\frac{1}{1-\frac{x}{a_n}}\Bigg]$$ Now, take the $\color{red}{\text{limit}}$ of the rhs when $x\to a_1$. You will ...


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