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Suppose I have;

$$f(x) = x^2 - 42x + 364$$

$$g(y) = y^2 - 35y + 364$$

Computing out values I find they have a common value

$$f(6) = g(8) = g(27) = f(36) = 148$$

Is there a way of finding these values?

Working through based on received help...

(1) $$f(x) = g(y)$$ (2) $$f(x) - g(y) = x^2 - 42x + 364 - y^2 + 35y - 364 = 0$$ (3) $$x^2 - 42x - y^2 + 35y = 0$$

Complete the squares...

(4) $$(x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4}$$ (5) $$4 ((x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4} )$$ (6) $$(4x^2 - 168x + 42^2) - (4y^2 - 140y + 35^2) = 42^2 - 35^2$$ (7) $$(2x - 42)^2 - (2y - 35)^2 = 539$$

Considering (7) as Fermat's factorization method;

(8) $$a^2 - b^2 = c$$ (9) $$(a + b)(a - b) = 539$$ (10) $$(2x - 42 + 2y - 35)(2x - 42 - 2y + 35) = 539$$ (11) $$(2x + 2y - 77)(2x - 2y - 7) = 539$$

Then, considering (11) as a simple product of two terms gives;

(12) $$d * e = 539$$ (13) $$(2x - 2y - 7) = d$$ (14) $$(2x + 2y - 77) = 539/d$$

CAB
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  • Lines! I don't see lines here. I think you are searching for the values $x$s in which $f(x)=g(x)$. So put $f(x)=g(x)$ and find the possible $x$ (s). – Mikasa Mar 24 '20 at 18:27
  • Curves, not lines. – Eric Towers Mar 24 '20 at 18:29
  • @mrs: For several hundred years before 1900 (or even the early 1900s), curves were usually called "lines". But I'm guessing CAB didn't time-travel here from the 1800s or is a long-lived vampire getting back into looking at math after a 150 year absence, so your initial reaction is the same one I had! – Dave L. Renfro Mar 24 '20 at 18:30
  • The curves given by $f$ and $g$ intersect, at $x = 0$. So "non-intersecting" in the title seems out of place. – Eric Towers Mar 24 '20 at 18:32
  • @DaveL.Renfro: My hands are up! Sorry... – Mikasa Mar 24 '20 at 18:36

3 Answers3

1

To find the common values $(x,y)$, for which $f(x)=g(y)$ you have to equate the two functions, and obtain: $$f(x)=g(y)\leftrightarrow x^2-42x+364=y^2-35y+364\leftrightarrow x^2-y^2-42x+35y=0$$ In the left sides, there is the equation of an hyperbola. Obviously, if $(x,y) \in R$, then there are infinite many solutions.

Matteo
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I presume you're looking for integers $x$ and $y$ such that $f(x) = g(y)$. By completing the square, the equation $f(x) = g(y)$ can be written as $$ (2x-42)^2 - (2y-35)^2 = (2x + 2y - 77)(2x - 2y -7) = 539$$ If $d$ is a (positive or negative) integer that divides $539$, you can solve $$ \eqalign{2x + 2y - 77 &= d\cr 2x - 2y - 7 &= 539/d\cr}$$ to get $$ \eqalign{x &= \frac{d^2 + 84 d + 539}{4d}\cr y &= \frac{d^2 + 70 d - 539}{4d}\cr}$$ and this will give all the solutions.

Robert Israel
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0

As the abovementioned 2 functions are continuous, you can find infinitely many common values in the intersection of their ranges.

VIVID
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