# find all solutions of (a,b)

#### Albert

##### Well-known member
$a,b$ are all integers

(1) $if : \,\, 5a^2+5ab+5b^2=7a+14b$

please find all solutions of $(a,b)$

#### MarkFL

##### Administrator
Staff member
My solution:

Applying the quadratic formula, we find:

$$\displaystyle b=\frac{(14-5a)\pm\sqrt{14^2-3(5a)^2}}{10}$$

Now we require:

$$\displaystyle 14^2-3(5a)^2\ge0$$

which implies:

$$\displaystyle a\in\{-1,0,1\}$$

Case 1: $$\displaystyle a=-1$$

$$\displaystyle b=\frac{19\pm11}{10}$$

The integral value is $$\displaystyle b=3$$

Hence, $(a,b)=(-1,3)$ is a solution.

Case 2: $$\displaystyle a=0$$

$$\displaystyle b=\frac{14\pm14}{10}$$

The integral value is $$\displaystyle b=0$$

Hence, $(a,b)=(0,0)$ is a solution.

Case 3: $$\displaystyle a=1$$

$$\displaystyle b=\frac{9\pm11}{10}$$

The integral value is $$\displaystyle b=2$$

Hence, $(a,b)=(1,2)$ is a solution.

#### Albert

##### Well-known member
My solution:

Applying the quadratic formula, we find:

$$\displaystyle b=\frac{(14-5a)\pm\sqrt{14^2-3(5a)^2}}{10}$$

Now we require:

$$\displaystyle 14^2-3(5a)^2\ge0$$

which implies:

$$\displaystyle a\in\{-1,0,1\}$$

Case 1: $$\displaystyle a=-1$$

$$\displaystyle b=\frac{19\pm11}{10}$$

The integral value is $$\displaystyle b=3$$

Hence, $(a,b)=(-1,3)$ is a solution.

Case 2: $$\displaystyle a=0$$

$$\displaystyle b=\frac{14\pm14}{10}$$

The integral value is $$\displaystyle b=0$$

Hence, $(a,b)=(0,0)$ is a solution.

Case 3: $$\displaystyle a=1$$

$$\displaystyle b=\frac{9\pm11}{10}$$

The integral value is $$\displaystyle b=2$$

Hence, $(a,b)=(1,2)$ is a solution.
thanks ,your answer is quite right

#### Albert

##### Well-known member
my solution:
let x=2a+b,and y=a+2b
$\therefore a=\dfrac {2x-y}{3},\,\,b=\dfrac {2y-x}{3}$
$5(a^2+ab+b^2)=7a+14b,\,\,becomes:$
$5(x^2-xy+y^2)=21y$
and we get :$y=5---(1),\,\, x^2-xy+y^2=21---(2)$
from (1)(2) (x,y)=(1,5)and (4,5)
and the corresponding (a,b)=(-1,3) and (1,2)
of course the third solution of (a,b)=(0,0)

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#### kaliprasad

##### Well-known member
my solution:
let x=2a+b,and y=a+2b
$\therefore a=\dfrac {2x-y}{3},\,\,y=\dfrac {2y-x}{3}$
$5(a^2+ab+b^2)=7a+14b,\,\,becomes:$
$5(x^2-xy+y^2)=21y$
and we get :$y=5---(1),\,\, x^2-xy+y^2=21---(2)$
from (1)(2) (x,y)=(1,5)and (4,5)
and the corresponding (a,b)=(-1,3) and (1,2)
of course the third solution of (a,b)=(0,0)
not correct as below
$5(x^2−xy+y^2)=21y$
and we get :y=5−−−(1),$x^2−xy+y^2=21$−−−(2)

is not strictly correct

it could be

y=5k−−−(1),$x^2−xy+y^2=21k$−−−(2)

#### Albert

##### Well-known member
not correct as below
$5(x^2−xy+y^2)=21y$
and we get :y=5−−−(1),$x^2−xy+y^2=21$−−−(2)

is not strictly correct

it could be

y=5k−−−(1),$x^2−xy+y^2=21k$−−−(2)
$5(x^2−xy+y^2)=21y----(@)$
we get :$21y\geq 5xy$ $(AP\geq GP)$
$\therefore x\leq 4$ (x,y being integers)
if x=4 ,y=5k
from (@) :$16-20k+25k^2=21k$
$16+25k^2=41k$
k must =1

#### kaliprasad

##### Well-known member
$5(x^2−xy+y^2)=21y----(@)$
we get :$21y\geq 5xy$ $(AP\geq GP)$
$\therefore x\leq 4$ (x,y being integers)
if x=4 ,y=5k
from (@) :$16-20k+25k^2=21k$
$16+25k^2=41k$
k must =1
could you explain me how

we get :21y≥5xy (AM≥GM) AM and GM of what ?

I an sorry that I could not understand

#### Albert

##### Well-known member
could you explain me how

we get :21y≥5xy (AM≥GM) AM and GM of what ?

I an sorry that I could not understand
$21y=5(x^2-xy+y^2)\geq 5(2xy-xy)=5xy$

$for: \,\, x^2+y^2\geq 2xy \,\, (AM\geq GM)$

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