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Why is my answer wrong? Consider the function on the integers given by \(f(x,y)=x^2y\). How many ordered pairs of integers satisfying \(−16\le x,y\le16\) is \(f(x,y)=f(y,x)\)?

Mathematics
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\[f(x,y)=f(y,x) \Rightarrow y^2x = x^2 y \]I observed that \(x\) and \(y\) cannot be negative here. So I am left with numbers ranging from \(0\) to \(16\). That gives me \(17\).
Are there any other ordered pairs where \(x \ne y\)?
ZOMG, wait.

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Other answers:

\((-4)^2 \times -4 = (-4)^2 \times -4\)
Facepalm. They can be negative.
PING and the light turns on!
:D
lol yeah. That fetches me \(17 + 16 = 33\)
Is 33 correct?
no
Aww.
you have 33+16+16=?
Why is that so?
that many ordered pairs (0,?) and (?,0)
Oh, I'm stupid. Thanks man!
so, it'd be 33+32+32 ordered pairs
how many ways can you have 1) x=0 2) y=0 3) x=y
(1,1),(2,2)...(16,16) is 16. (-1,-1)...(-16,-16) is 16. (0,?) is 33. (?,0) is 33. 32 + 33 + 33 = 98 We overcounted a (0,0), so 32 + 32 + 33 = 97. Right?
Is this from brilliant.org?
@oldrin.bataku Yeah, practicing stuff :-)
this way you counted (0,0) twice
Am I not allowed to ask questions for practice?
@electrokid I subtracted it in the end.
right.
I'm not sure if you're allowed to ask for help on brilliant.org questions, at least publicly... :-p
yes you are allowed
and i am god
@oldrin.bataku But I think I'm just practicing techniques.
brilliant! 97 is the correct answer
o'course it'd be lol
u wbhi dheu dhee hee:)
@perl what?
i said, i enjoyed the problem
now i must go , there are babies to punish

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