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# Equations

0
128
1

x+\sqrt y = 27

y +\sqrt x = 125
Given that x and y are positive integers satisfying the system of equations above, find x+y.

Jun 1, 2022

#1
+1

Hello Guest!

$$x+\sqrt{y}=27$$      Equation (1)

$$y+\sqrt{x}=125$$    Equation (2)

x and y must be perfect squares, as we are taking their square roots.

So let: $$x=n^2$$ and $$y=m^2$$ $$\text{for some positive integers n,m }$$

Then, Equation (1) and Equation (2) becomes:

$$n^2+m=27$$           (*)

$$m^2+n=125$$         (**)

Notice, from equation (2):

$$y<125 \implies m^2 \le 121 \implies m\le11$$  (Note: y is a perfect square, so m must be an integer, there is no square root for 125 in integers, then we take the next least square root, that is 121, hence we deduce the above inequalities).

Also: $$x<27 \implies n^2\le25 \implies n\le 5$$ (Similarly , there is no square root of 27 in integers, so we take the next least square root, which is 25).

Now, from (**) Isolate n and we know $$n\le 5$$

Thus,  $$n=125-m^2 \le5$$

Focus on: $$125-m^2\le5$$ and solve this inequality for m. (Remember, m is an integer, so if the square root of 120 does not exist in integers, we take the next least or greater square root, depending on the inequality sign as below.)

$$120 \le m^2 \implies m\ge 11$$  But, we know $$m\le 11$$ from equation (2), and now we deduced $$m\ge11$$. Therefore, we completely bounded m, that is, m must be 11 to satisfy both inequalities.

Well, since: $$m=11$$, then from (*):

$$n^2=16 \implies n=4$$ Also from (**): $$n=125-11^2=125-121=4$$

So we found n and m.

Hence,

$$x=n^2 \implies x=4^2=16$$ and $$y=m^2 \implies y=11^2=121$$

Therefore, $$x+y=16+121=137$$

I hope this helped.

Jun 2, 2022
edited by Guest  Jun 2, 2022