Question 1208295
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If 2¹³ + 2¹⁰ + 2ˣ = y², find x and y.
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        It is assumed that  x  and  y  are integer numbers.


        I will give another,  strict mathematical solution.



<pre>
{{{2^13}}} + {{{2^10}}} = 9216 = 96^2.       (1)


Therefore,  {{{2^x}}} = {{{y^2}}} - {{{96^2}}}.    (2)


Hence,  {{{2^x}}} = (y+96)*(y-96).      (3)


Thus,  {{{2^x}}} is the product of integers y+96 and y-96.


From it, we conclude that  y+96 and y-96 are degrees of 2.

So, we write

    y + 96 = {{{2^n}}}

    y - 96 = {{{2^m}}}


with integer non-negative n and m, and we understand that m < n.  
Subtracting the lower equation from the upper one, we get

    {{{2^n}}} - {{{2^m}}} = 192,

     {{{2^m*(2^(n-m)-1)}}} = 192 = 64*3 = {{{2^6*3}}}.   (4)


From it, we conclude that  m = 6, n-m = 2;  hence n-6 = 2,  n = 8.


Thus    y+96 = {{{2^n}}} = {{{2^8}}} = 256;  hence  y = 256-96 = 160.

<U>CHECK</U>:  y-96 = {{{2^m}}} = {{{2^6}}} = 64;  hence  y = 64+96 =160.   <<<---===  Check says OK.


Now  from (3)  

    {{{2^x}}} = (y+96)*(y-96) = (160+96)*(160-96) = 16384 = {{{2^14}}},


Hence,  x = 14.


<U>ANSWER</U>.  This problem has two solutions  (x,y) = (14,160)  and  (x,y) = (14,-160).


From where the second, negative value of y came ?


Since right side of equation (1) is y^2, it is clear that with positive solution y= 160,
negative solution y= -160 works, too.


Where we missed it in our reasoning ? - Because in (4), we could take NEGATIVE factors  {{{-2^6}}}  and  -3 
into consideration.  It would lead us to the negative value of y.

But since we just caught this second solution, we shouldn't worry anymore.
</pre>

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Thus the problem is solved completely, using strict mathematical reasoning, 
and two solutions in integer numbers are found.  All necessary explanations are given.


Happy learning (!)