A Perfect Square

Because for the Perfect Spuare formula $(a+b)^{2}$ = $a^{2}+2ab+b^{2}$

So ,we need to make $2^{k}+2^{34}+2^{37}$ like a Perfect Square,

• $2^{k}+2^{34}+2^{37}$ = $[(2^{17})]^{2}+2(2^{17})(2^{19})+2^{k}$

Thus, $k =19 \times 2 = \boxed {38}$

The square of a natural number $n$ can be expressed as $n^{2}$ . We need to find the value of $k$ such that $2^{k}+2^{34}+2^{37}$ can be expressed as $n^{2}$ . $2^{k}+2^{34}+2^{37}$ $=2^{k}+2^{34}(1+2^{3})$ $=2^{k}+2^{34}(1+8)$ $=2^{k}+2^{34}(9)$ $=2^{34}(2^{k-34}+9)$ $=(2^{17})^{2}(2^{k-34}+9)$ Taking $x=2^{17}$ , we get, $=x^{2}(2^{k-34}+9)$ As, out of the two factors of $n^{2}$ , one can be expressed as a square, we simply need to find a value for $k-34$ such that $2^{k-34}+9$ can be expressed as a square. We know that, $16+9$ $=25$ can be expressed as a square i.e. $5^{2}$ . Thus,the value of $2^{k-34}=16$ $\Rightarrow 2^{k-34}=2^{4}$ $\Rightarrow k-34=4$ $\Rightarrow k=\boxed{38}$ Thus the value of $k$ is $\boxed{38}$ .

According to the theory, we have

$2^k+2^{34}+2^{37}=n^2 \Leftrightarrow 2^k=n^2-( 3.2^{17})^2$

$\Leftrightarrow 2^k=( n+3.2^{17} ) (n-3.2^{17} )$

Hence,

$\left\{ \begin{array}{l l} 2^a=n+3.2^{17} & \quad \text{}\\ 2^b=n-3.2^{17} & \quad \text{} \end{array} \right.$

in which $a;b\in \mathbb{N},k=a+b,a>b$

$\Rightarrow 2^a-2^b=3.2^{18} \Leftrightarrow 2^b(2^{a-b}-1)=3.2^{18}$

$2^{a-b}-1$ is odd so,

$\left\{ \begin{array}{l l} 2^{a-b}-1=3 & \quad \text{}\\ 2^b=2^{18} & \quad \text{} \end{array} \right.$

$\Rightarrow b=18;a=b+2=20$

Then $k=38.$

We know that 2^k+2^{34}+2^{37}=n^2 .....1

so taking 2^{34} common from the equation 1

so,we get 2^{34}(2^{k-34}+1+8)

as 2^34 is perfect square of 2^17 .so n to be perfect integer,

(2^{k-34}+9)...2 equation 2 should also be prefect square root

so to see value of square root , we put value k\geq34 so we put value of k in equation 2, k=34 we get 10 which is not a perfect square root : : : : k=38 we get 25 which is a perfect square of 5

therefore value of k is 38

if this is the square of an integer so we can write the integer on this form [x+y]^2 we have 2^[34] that we can sqrt it to 2^[17] So our integer will be on this form [x + 2^[17] ]^2 = 2^k +2^[34] +2^[37] Note that 2^[37] is not perfect Square, so 2^[37] = 2^[18] * x x = 2^[19] 2^k = x^2 So 2^[38] = 2^k

Considere o produto notável (x+y)^2 = x^2 + 2xy +y^2. Como 2^37 não é quadrado perfeito, faça com que ele seja o termo 2xy do produto notável e que 2^34 seja o x^2, entao y deve ser 2^19 e y^2 deve ser 2^38.

We have $\small 2^k+2^{34}+2^{37} = m^2 \implies \small 2^{34}(2^{k-34} + 9)=m^2$

Now $\small 2^{34}$ is a perfect square. This implies $(\small 2^{k-34} + 9)$ is a perfect square. So let $\small(2^{k-34} + 9) = p^2 \implies \small 2^{k-34}=(p-3)(p+3)$ Now the only powers of $2$ that differ by $6$ are $2$ and $8$ . So we finally have $\small 2^{k-34}=2^4 \implies \boxed {k=36}$

2^{34}+2^{37}+2^k=2^{34}[1+2^3+C]=2^{34}[9+C] If only C=16 then it will be perfect square. 16=2^4 So 2^{38}=2^k k=38

Let k be a natural number such that

2^k + 2^{34} + 2^{37} = n^2

where, n is any integer

2^k + 2^{34} X ( 1 + 2^3 ) = n^2

2^k + ( 9 X 2^{34} ) = n^2

2^k + ( 3 X 2^{17} )^2 = n^2

2^k = n^2 - ( 3 X 2^{17} )^2

2^k = ( n + 3 X 2^{17} ) X ( n - 3 X 2^{17} )

Since, the right hand side must be having no prime other than 2 in its factorization, hence, we have to get rid of the '3' present. This can be done of n is of the form ( 2m - 1) X 2^{17} as odd - odd = even.

The only 'm' satisfying this is m = 2 as for m=0,1 the expression will be negative, which is not possible while for m>2 other primes addition to 2 will come.

Hence, after putting n = 5 X 2^{17} in the equation,

2^k = ( 8 X 2^{17} ) X ( 2 X 2^{17} )

2^k = 2^{38}

=> k = 38

Let $a_0$ and $b_0$ be numbers such that $a_0 + b_0 = n$ . Now we want $(a_0 + b_0)^{2} = a_0^{2} + b_0^{2} + 2 \times a_0 \times b_0 = 2^{k} + 2^{34} + 2^{37}$ It is easy to see that $a_0^{2} = 2^{k}$ , $b_0^{2} = 2^{34}$ and $2 \times a_0 \times b_0 = 2^{37}$ . So $2 \times a_0 \times b_0 = 2^{37} = 2 \times 2^{k/2} \times 2^{34/2} = 2^{\frac{k}{2} + 18} = 2^{37}$ Therefore $\frac{k}{2} + 18 = 37$ and then $k = 38$

$2^{k}+2^{34}+2^{37}=2^{k}+2^{34}(1+2^{3})=2^{k}+2^{34}(3^{2})$

Assuming that $k>34$ , $2^{k}+2^{34}(3^{2})=2^{34}(2^{k-34})+2^{34}(3^{2})$

Since $34$ is even, we can treat this as a pythagorean triple where $(2^{17}\cdot5)^{2}=(2^{17}\cdot4)^{2}+(2^{17}\cdot3)^{2}=2^{34}(2^{k-34})+2^{34}(3^{2})$

Now, by letting $2^{34}(2^{k-34})=(2^{17}\cdot4)^{2}$ and solving for $k$ , we get $\boxed{k=38}$

$2^{k} + 2^{34} + 2^{37}$

should be a perfect square

This equation is equal to

$2^{34} (2^{k-34} + 2^{0} + 2 ^{3})$

$2^{34} (2^{k-34} +1 + 8)$

since $2^{34}$ is already a perfect square

$(2^{k-34} + 1 + 8)$ should be a perfect square too

$(2^{k-34} + 9 = 25)$

$(2^{4} + 9 = 25)$

so $k-34 = 4$

$k=38$

$2^k+2^{34}+2^{37}=2^34\times (9+2^{k-34})$ and $16+9=25=5^2$ . But here we have only found the smallest such $k$ , there may be other values of $k$ such that $2^k+2^{34}+2^{37}$ is a perfect square.

<=> 2^k + 2^34 + 2^37 = 2^k + 2^34.(1 + 2^3) = n^2 <=> 2^k + 2^34 . 9 = n^2 <=> Because 2^34 and 9 is perfect square, so 2^k also is perfect square. We know if 25 = 9 + 16 and 2^34 become distributor number. So, 2^k = 2^p . 16 = 2^34. 16 => 2^k = 2^(34+4) = 2^38. Then, we have k = 38

2^{k}+2^{34}+2^{37}= 2^{34}(2^{k-34}+1+8) 2^{34} is an perfect square 2^{k-34}+9 should also perfect square k=38

N= 2^34(2^(k-34)+1+8) taking 2^34 common N=2^34(2^(k-34)+9) so for N to be a perfect square 2^(k-34)+9 should be perfect square for 2^(k-34)+9=25 k=38