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Source: STEMS 2025 Category B4, C3

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113 posts

#1 h Dec 29, 2024, 12:35 PM • 1 Y

Y by starchan

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$ , $xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)$ Proposed by Aritra Mondal

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#2 h Dec 29, 2024, 1:06 PM • 2 Y

Y by HotSinglesInYourArea, Calamarul

Swapping $x$ and $y$ we get $xf(x+y)+(x+y)f(y)=yf(x+y)+(x+y)f(x)\iff (x+y)(f(x)-f(y))=(x-y)f(x+y)$ , which has been posted here:
https://artofproblemsolving.com/community/q1h365241p2008230

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254 posts

#3 h Dec 29, 2024, 1:25 PM • 1 Y

Y by Om245

The solutions are of the form $f(x) = cx$ . $P(x, y) - P(y, x)$ gives us $\frac{f(x+y)}{x+y} = \frac{f(x)-f(y)}{x-y}$ and the function can be seen to be odd. Now consider the assertions $P(1+x, -x)$ , $P(-x, x-1)$ and $P(x-1, 1+x)$ . These give us that $f(x) = \left(\frac{x^2+x}{2} \right) f(1) + \left(\frac{x^2-x}{2}\right) f(-1) = f(1)x$ which can be seen to work. $\blacksquare$

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163 posts

#5 h Mar 10, 2025, 6:13 PM

Y by

Double07 wrote:

Swapping $x$ and $y$ we get $xf(x+y)+(x+y)f(y)=yf(x+y)+(x+y)f(x)\iff (x+y)(f(x)-f(y))=(x-y)f(x+y)$ , which has been posted here:
https://artofproblemsolving.com/community/q1h365241p2008230

i dont think the answer can be a quadratic :noo:

:noo:

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453 posts

#6 h Mar 20, 2025, 9:03 AM

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Uhm, the coefficient of the quadratic term $a$ can be $0$ .

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#7 h Mar 20, 2025, 9:50 AM

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SatisfiedMagma wrote:

Uhm, the coefficient of the quadratic term $a$ can be $0$ .

that's just saying it's $cx$ , the $ax^{2}$ would be redundant, no use of that then why mention it

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#8 h Mar 22, 2025, 11:13 PM

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Levieee wrote:

SatisfiedMagma wrote:

Uhm, the coefficient of the quadratic term $a$ can be $0$ .

that's just saying it's $cx$ , the $ax^{2}$ would be redundant, no use of that then why mention it

Call the original STEMS FE to be $(\star)$ and referenced FE to be $(\clubsuit)$ . Now since $(\star) \implies (\clubsuit)$ , if $f$ is a solution, then $f$ is a solution for $(\clubsuit)$ as well. But if you analyze $(\clubsuit)$ , then you deduce that that quadratic polynomials work with certain constraints work for $(\clubsuit)$ . But remember, this doesn't mean that $f$ must be a solution to the equation $(\star)$ . Thefore, you gotta check and put it back in $(\star)$ . If you do the putting, then you get $a = 0$ . The thing is OP, wanted to say, you can reduce the problem to a known-problem(which was different than the STEMS problem here, that's why he wrote a quadratic term in the answer, because he was solving a different problem).

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