How to prove that if $x^2+c\sqrt{b+x^2}=y^2+c\sqrt{b+y^2}$ then $x=y$?

Question

I have a function of $x^2$ as $f(x^2)=x^2+c\sqrt{b+x^2}$ and $f(y^2)=y^2+c\sqrt{b+y^2}$ now suppose $f(x^2)=f(y^2)$, How can I prove that $x=y$ if there is a constraint that both $x$ and $y$ are positive integers. Taking squares do not help as they further add more terms with square roots.

Can I show that show using graph? But the graphs will overlap.

Answer 1

$$ \begin{aligned} & f\left(x^2\right)=x^2+c \sqrt{b+x^2} \\ & f\left(y^2\right)=y^2+c \sqrt{b+y^2} \end{aligned} $$

x &y are positive integers $$ \text { if } f\left(x^2\right)=f\left(y^2\right) \text { prove that } x=y $$ Let's assume, x not equal to y.

Let $\text{ Case 1: if x<y}\Rightarrow y=x+h, \Rightarrow h \in \mathbb{Z}^{+}$ $$ \begin{aligned} \therefore f\left(y^2\right) & =(x+h)^2+c \sqrt{b+x^2+h^2+2 x h} \\ & =x^2+h^2+2 x h+c \sqrt{b+x^2+h^2+2 x h} \\ & =f\left(x^2\right)=x^2+c \sqrt{b+x^2} \end{aligned} $$ $\Rightarrow h^2+2 x h=c\left\{\sqrt{b+x^2}-\sqrt{b+x^2+h^2+2 x h}\right\}$

Since, x is a positive integer, we can clearly see that no such $h$ exists.

Case 2: if $x>y \Rightarrow x=y+h, \Rightarrow h \in \mathbb{Z}^{+}$ $$ \begin{aligned} f\left(x^2\right) & =f\left(y^2\right) \\ \Rightarrow(y+h)^2+c \sqrt{b+(y+h)^2} & \\ & =y^2+c \sqrt{b+y^2} \\ \Rightarrow h^2+2 y h & =c\left\{\sqrt{b+y^2}-\sqrt{b+(y+h)^2}\right\} \end{aligned} $$

Again its clear no such $h$ exists. Why? Since y is a positive integer.

$\therefore$ our assumption falls, Hence $x=y$, Q.E.D.

Answer 2

If $\;b\ge0,c>0,$ the function $$f(t)=t+c\sqrt{b+t}$$ is the sum of two increasing functions on $[0,\infty),$ so it is increasing on $[0,\infty).$ Therefore, $f$ is injective and $$f(x^2)=f(y^2) \quad \iff \quad x^2=y^2.$$ The claim is true for any positive reals $x,y,$ not only for integers.

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If $\;b\ge0, c<0,$ the function is defined on $[0,\infty),$ but is not monotone. The claim is false, as shows the following counterexample:

Let $b=5, c=-3-\sqrt6.$

Then $$f(1^2)=1-(3+\sqrt6)\sqrt{6}=-5-3\sqrt6$$ and also $$f(2^2)=4-(3+\sqrt6)\sqrt{9}=-5-3\sqrt6$$ For the integers $x=1, y=2$ are the values equal, but $x\neq y.$