Question

$\: \: \large { \underline {\underline{\bf{Question}} \: \: }}_{ \bigstar \star}$
$\odot \: \: \: \: \sf{if \: \: 615 + {x}^{2} = {2}^{y} \: \: then \: \: find \: \: the \: \: values \: \: of \: \: x \: \: and \: \: y }\bigg[ \bf x,y \in \Z\bigg]$
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Answer 1

Topic :-

Number Theory

Given :-

$\mathtt{615+x^2=2^y}$

To Find :-

$\mathtt{Value\:of\:'x'\:and\:'y'\:where\:x,\:y \:\epsilon\:\mathbb{Z}}$

Solution :-

$\mathtt {615\:is\:an\:odd\:number.}$

$\mathtt {2^y\:is\:an\:even\:number.}$

Odd + x² = Even

x² = Even - Odd

x² = Odd

Square of an odd number is odd.

So, x = Odd.

Possible last digit of 'x' = 1, 3, 5, 7, 9.

Possible last digit of x² = 1, 9 and 5.

$\mathtt{Possible\:last\:digit\:of\:2^y = 2,\:4,\:8\:and\:6.}$

but adding a number with last digit 1, 9 and 5 to 615 will give a number with last digit as 6, 4 and 0.

So, only possible last digit of $\mathtt{2^y}$ are 4 and 6.

$\mathtt{2^y}$ ends with 4 and 6 only when 'y' is even.

We can write y = 2t.

$\mathtt{615+x^2=2^y}$

$\mathtt{2^{2t}-x^2=615}$

$\mathtt{(2^{t}+x)(2^{t}-x)=615}$

Sum of factors,

$\mathtt{(2^{t}+x)+(2^{t}-x)}$

$\mathtt{2(2^t)=2^{t+1}}$

Possible factors of 615,

1 × 615

3 × 205

5 × 123

15 × 41

Sum of possible factors,

1 + 615 = 616

3 + 205 = 208

5 + 123 = 128

15 + 41 = 56

The only Sum which is in power of 2 is 128.

$\mathtt{2^{t+1}=128}$

$\mathtt{2^{t+1}=2^7}$

$\mathtt{t+1=7}$

$\mathtt{t=6}$

So, y = 2t = 2(6) = 12

Calculating value of 'x',

Put value of y in original equation to get value of 'x'.

$\mathtt{615+x^2=2^y}$

$\mathtt{615+x^2=2^{12}}$

$\mathtt{x^2=4096-615}$

$\mathtt{x^2=3481}$

$\mathtt {x=59,-59}$

Answer :-

So, value of y is 12 and value of x are 59 and -59.