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[723] juantheron

2012-05-18 07:06:44

Thanks friends got it.

New task::

Numerical value of cos (1⁰).cos (3⁰).cos (5⁰).......cos (89⁰)=

where all angle are in Degree

[722] m2mm

2012-05-17 21:16:33

We may assume that x $\le$ y. Then (y+2)²=y²+4y+4>y²+3x>y². So, if y²+3x is a square, it must be (y+1)². Then y²+2y+1=y²+3x, so 2y+1=3x, which means that 2y $\equiv$ -1 (mod 3), so y $\equiv$ 1 (mod 3). Hence, y=3k+1 for some positive integer k. Since 2y+1=3x, then 6k+3=3x, 2k+1=x.

As x²+3y is a square, then (2k+1)²+3(3k+1)=4k²+4k+1+9k+3=4k²+13k+4 as well.

But (2k+2)²=4k²+8k+4 $\le$ 4k²+13k+4<4k²+16k+16=(2k+4)², so 4k²+13k+4=(2k+2)²=4k²+8k+4, or 4k²+13k+4=(2k+3)².

If 4k²+13k+4=4k²+8k+4, then k=0, x=y=1 (and this pair is a soultion).

If 4k²+13k+4=(2k+3)², then 4k²+13k+4=4k²+12k+9, so k=5. Then the solution is (11,16) or (16,11).

Előzmény: [717] juantheron, 2012-05-16 20:17:02

[721] logarlécész	2012-05-17 20:25:55
Teljesen igaz. Hiába, a figyelmetlenség mindig játszik. :-)
Előzmény: [712] nadorp, 2012-05-16 14:16:10

[720] nadorp

2012-05-17 14:18:36

"(x-y).(x+y-3)=(l+m).(l-m) means either x+y-3=l+m and x-y=l-m "

Incorrect. For example

3^.8=6^.4 but 8 $\neq$ 6 and 3 $\neq$ 4

Előzmény: [719] juantheron, 2012-05-17 08:49:02

[719] juantheron

2012-05-17 08:49:02

Thanks Nadorp got it

task:: find positive Integer value of x and y for which x²+3y and y²+3x is an perfect square

My solution::

Let x²+3y=l² and y²+3x=m²

Where l,m $\in$ Z

Then (x²-y²)-3(x-y)=l²-m²

(x-y).(x+y-3)=(l+m).(l-m)

means either x+y-3=l+m and x-y=l-m

2x=3+2l or $x=\frac{3+2l}{2}\notin Z$

similarly $y=\frac{3+2m}{2}\notin Z$

but when i put rendomly put values like (1,1) and (11,16) and (16,11)

but how can i calculate it

Thanks

[718] nadorp

2012-05-17 08:23:48

"But Not sure about k7 has no value for which satisfy the given equation"

Answer:

x²=4+[x] $\leq$ 4+x

x²-x-4 $\leq$ 0

$-\sqrt3<\frac{1-\sqrt{17}}2\leq x\leq\frac{1+\sqrt{17}}2<\sqrt7$

Előzmény: [716] juantheron, 2012-05-16 19:46:00

[717] juantheron

2012-05-16 20:17:02

Nwe task::

The number of ordered pairs of positive integers x,y such that x²+3y and y²+3x are both perfect squares

[716] juantheron

2012-05-16 19:46:00

Yes Friends your answer is Right

I have solved like that way

x²=4+[x]

here R.H.S(right hand side) is an Integer that means L.H.S(Left hand side must be an Integer)

So x² $\in$ Z

So Let x²=k, where k>0

So $x=\pm \sqrt{k}$

So our equation convert into $k=4+[\pm \sqrt{k}]$

Now after checking certain value of k like k=1,2,3,4,5,6

We Get only k=2 and k=6 satisfy the given equation

and after k $\geq$ 7, we get L.H.S # R.H.S

But Not sure about k $\geq$ 7 has no value for which satisfy the given equation

can anyone clear last line explanation

Thanks

[715] juantheron

2012-05-16 19:35:38

Thanks Nadorp and Small Potato

for Integration I have solve it after yours work

I have solved it like in that way

$\int \frac{1}{\sin^3 x+\cos^3 x}dx$

$\int\frac{\sin^2 x+\cos^2 x}{(\sin^3 x+\cos^3x)}dx$

$=\frac{1}{3}(\int\frac{\sin x+\cos x}{\sin^2 x-\sin x.\cos x+\cos^2 x}+\int\frac{2}{\sin x+\cos x}dx$

$-\frac{2}{3}\int \frac{(\sin x+\cos x)}{1+(\cos x-\sin x)^2}dx+\frac{2}{3}\int\frac{1}{\sin x+\cos x}dx$

Put (cos x-sin x)=tand (sin x+cos x)dx=-dt

and $\sin x+\cos x=\sqrt{2}.\sin (x+\frac{\pi}{4})$

after that we can calculate easily

[714] SmallPotato	2012-05-16 14:21:38
Lekéstem :-)
Előzmény: [712] nadorp, 2012-05-16 14:16:10

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