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$\text{integer number }x,y >0\\f(x) = x^2 + 4\\ f(y) = x^2 + 23\\ f(x-y) = ?$

My Work:
This doesn't make any scene to me..
The second equation is telling that
$f(y)=x^2 + 23$ for any positive integer $y$. If we set $y=x$ then $f(x) = x^2 + 23$
This would imply that
$x^2 + 4 = x^2 + 23 \implies 4 = 23$ !!

I know that I am wrong somewhere.... Or I didn't understand the problem well.. Any Hint will be helpful :)

PS: This is a problem from BdMO 2006 Regionals

Thomas
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Rezwan Arefin
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  • Is it not a typo when you have written $f(y)=x^2+23$ instead of $f(y)=y^2+23$?. This would be the reason of you are wrong. – Piquito Sep 30 '16 at 22:36
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    $f(y) = x^2 + 23$ makes sense only if y = g(x). which .... it very well could. – fleablood Sep 30 '16 at 22:36
  • @SimpleArt BdMO - Bangladesh Math Olympiad... I think you don't know bengali ... – Rezwan Arefin Sep 30 '16 at 22:38
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    No, you are given numbers $x$ and $y$, and the equations follow from what is given. – abiessu Sep 30 '16 at 22:39
  • If $x=1, y=2$ then the equations only tell you that $f(1)=5$, $f(2)=24$. So then $f(x-y)=f(-1)$ can be anything. So, likely one of those equations must hold "for all $x$" or "for all $y$" while the other holds only for a specific $x$ or $y$. Or there is some other typo. Overall, this question needs clarification. – Michael Sep 30 '16 at 22:44
  • @abiessu that wasn't given until a later edit of this post. – fleablood Sep 30 '16 at 23:27
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    @RezwanArefin The question is incomplete and can not be answered in its present form. Some BdMO downloads appear to have English translations. Does the problem come from one of those? – dxiv Sep 30 '16 at 23:38
  • @fleablood: I definitely agree with the ambiguity. – abiessu Oct 01 '16 at 01:59

3 Answers3

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If $f(x) = x^{2}+4$, then $f(y) = y^{2}+4 = x^{2}+23$, so $y = \sqrt{x^{2}+19}$. Then $$f(x-y) = (x-y)^{2}+4 = x^{2}-2xy+y^{2} + 4 = x^{2}-2xy + x^{2}+23 = 2x^{2}-2x\sqrt{x^{2}+19} +23$$

xxxxxxxxx
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    This interprets the problem as being "$f(x) = x^2 + 4$ for all positive integers $x$," but the equation $f(y)=x^2+23$ as being true only for a particular positive integer $y$ (and then, for what value of $x$ does it hold?). This is not clear from the problem statement. I think it is just as likely there is a typo somewhere, or some other info is missing. – Michael Sep 30 '16 at 22:57
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    @Michael Even if $f(x) = x^2 + 4$ for all positive integers $x$, that's still not enough to calculate $f(x-y)$ where the argument $x - y = x - \sqrt{x^2 + 19}$ is not necessarily an integer. The problem is badly mis-stated, and there is simply not enough information to even guess what the real question was. – dxiv Sep 30 '16 at 23:33
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The key is the line $x, y \ge 0$. That are not variables but specific values. If it helps use $a, b \ge 0$.

Let $f:\mathbb R \rightarrow \mathbb R$. $f(x) = ....\text{we have no idea... they didn't tell us anything}$

$f(a) = a^2 + 4$ (Not a definition! A coincidence. For example: $f(x) = 4x; a = 2; f(a) = 4*2 = 8 = 2^2 + 4= a^2 + 4$. Just happens to be true for $a$. Not for any other numbers in general.

$f(b) = a^2 + 23$. I dunno. $f(b) = f(a) + 19$... don't see any good that'll do.

What is $f(a-b)= ?$.

Don't know. Could be anything.

Say $f(x) = x + 6$ then $f(a=2) = 8 = a^2 + 4$ and $f(b=21) = 27 = a^2 + 23$ so $f(a - b) = f(-19)= -19 + 6 = -13$.

But if $f(x) = 19x - 14; a = 1; f(a) =5 = 1^2 + 4;b= 2; f(b)=24 = 1^2 + 23$ but $f(a-b) = f(-1) = -19-14 = -33$.

Pointless, I think.

fleablood
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  • I think this answer agrees with my comment above. I do not know why there are other answers, and why there are actually "plus 1" votes for them. – Michael Sep 30 '16 at 23:13
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    Actually my first interpretation was Simple Art's interpretation below. But I didn't like that the question didn't specify that f(x) =x^4 + 4 was a definition while y(x) was a function . And then the edit x , y positive integers throws that interpretation out the window. So upshot... no freaking idea what the question actually means. – fleablood Sep 30 '16 at 23:20
  • None of the given answers even specify an interpretation. The question itself re-uses $x$ as a variable in both equations. It does not make sense to assume one of the equations holds "for all $x$" and the other holds for some specific value of $x$. Anyway, I didn't downvote any answers, even though they add to the ambiguity. – Michael Sep 30 '16 at 23:33
  • The one thing you can NOT do is assume f(x):= x^2 + 4 is definition and then assume it is a variable "solve for x and y f(y) = y^2 +4 = x^2 +23" in the second. But I think y_x so that f(y_x) = y_x^2 + 4 = x^2 + 23 so y = \pm \sqrt{x^ + 19} is fair but if so the question should INDICATED that was the interpretation. – fleablood Sep 30 '16 at 23:45
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Hint:

Assume $y=g(x)$. We want to find a $g(x)$ that satisfies

$$f(g(x))=x^2+23$$

$$g(x)=f^{-1}(x^2+23)$$

$$y>0\tag{given}$$

$$f^{-1}(x)=\pm\sqrt{x-4}$$

$$\boxed{y=g(x)=\sqrt{x^2+19}}$$

Simply Beautiful Art
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    I'm upvoting this because it is a possible interpretation that would make sense. However I think the problem is really remiss for not specifying it as such. It could jut as well be interpreted as x and y are specific numbers ant there is no solvable solution. – fleablood Sep 30 '16 at 23:30