Two problems from 2012 India IMO Training Camp

Let $a\geq b$ and $c\geq d$ be real numbers. Prove that the equation \[(x+a)(x+d)+(x+b)(x+c)=0\] has real roots.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y+xy)=f(x)+f(y)+f(xy) \ \forall x,y \in \mathbb{R}$ . Prove that $f$ satisfies $f(x+y)=f(x)+f(y) \ \forall x,y \in \mathbb{R}$ .

#Algebra

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I'll post my solutions one day later in the comments (in this reply section) :)

ChengYiin Ong - 4 months ago

Let $f(x)=(x+a)(x+d)+(x+b)(x+c)$ , then $f(-a)=(b-a)(c-a)$ and $f(-c)=-(d-c)(c-a)$ . We have $f(-a)\cdot f(-c)=-(b-a)(d-c)(c-a)^2\le 0$ which means $f(-a)$ or $f(-c)$ is equal to zero which means there is a real root, or $f(-a)$ and $f(-c)$ have different signs which also tells us that $f$ has a real root by IVF.
First, set $x,y \rightarrow 0$ , we get $f(0)=0$ . Next, set $y\rightarrow 1$ , we have $f(2x+1)=2f(x)+f(1)$ . Then set $x\rightarrow uv+u+v , y\rightarrow 1$ , we get $f(2uv+2u+2v+1)=2f(uv+u+v)+f(1)=2f(u)+2f(v)+2f(uv)+f(1).$ We can compute $f(2uv+2u+2v+1)$ in another way, set $x\rightarrow 2u+1, y\rightarrow v$ we have $f(2uv+2u+2v+1)=f(2u+1+v+2uv+v)=f(2u+1)+f(v)+f(2uv+v)=2f(u)+f(v)+f(2uv+v)+f(1).$ Comparing these two results, we see that $f(2uv+v)=2f(uv)+f(v).$ By setting $u\rightarrow -\frac{1}{2}$ , we have $2f\left(-\frac{v}{2}\right)=-f(v) \implies 2f(-x)=-f(2x)$ for all real $x$ . So, $-f(-2uv)+f(v)=2f(uv)+f(v)=f(2uv+v)$ let $s=2uv$ and $t=v$ , we have $f(s+t)+f(-s)=f(t)$ and by changing variables again, let $x=-s, y=s+t$ , we finally arrive at $f(x)+f(y)=f(x+y). \quad \blacksquare$

ChengYiin Ong - 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$