An algebra problem by Shariful Islam

Note that:

$3^2+4^2=5^2$

$10^2+11^2+12^2=13^2+14^2$

$21^2+22^2+23^2+24^2=25^2+26^2+27^2$

$36^2+37^2+38^2+39^2+40^2=41^2+42^2+43^2+44^2$

An interesting exercise is to prove for any $n$ we can find a string of $n$ consecutive squares that are equal to the squares of the following $n-1$ numbers.

If you look at the L.H.S 10 will always give a unit digit (0) , 11 will always give (1) and 12 will give (2,4,8,6) and repeating. So the unit digit for the result will be (3,5,9,7).
If you look at the R.H.S 13 will give (3,9,7,1) and repeating , 14 will give (4,6,4,6) and repeating . So the unit digit for the result will be (7,5,1,7).
By comparing the results (3,5,9,7)
(7,5,1,7)
the answer is (2).

I guessed, i plugged in 1, then plugged in 2 then got it correct

Firstly, x=2 is a root.

Divide both sides to $13^x$ :

$(\dfrac{10}{13})^x+(\dfrac{11}{13})^x+(\dfrac{12}{13})^x=1+(\dfrac{14}{13})^x$

$\rightarrow (\dfrac{10}{13})^x+(\dfrac{11}{13})^x+(\dfrac{12}{13})^x-1-(\dfrac{14}{13})^x=0$ (1)

Let $f(x)$ equals the left side of (1). Then:

$f'(x)=-log(\dfrac{13}{10})(\dfrac{10}{13})^x-log(\dfrac{13}{11})(\dfrac{11}{13})^x-log(\dfrac{13}{12})(\dfrac{12}{13})^x-log(\dfrac{14}{13})(\dfrac{14}{13})^x$

Since $log(\dfrac{a}{b})>0$ with $a>b$ , $f'(x)<0$ with all real $x$ . Hence $f(x)$ have at most one root.

So answer is $\boxed{x=2}$