Equation With Integer Solutions

[anemone]

Solve the following equation for integer solutions:

$a^2(b-2)+b^2(a-2)+28=0$

 

[Albert1]

Solve the following equation for integer solutions:

$a^2(b-2)+b^2(a-2)+28=0---(A)$

Spoiler

suppose the original equation $A$ can be factorized as $(B)$ or $(C)$
$a^2(b-2)+b^2(a-2)+28=0----(A)\\
(a^2-x)[(b-2)-\dfrac{28}{x}]=0---(B)\\
\therefore x=1,\,\, or\,\, x=4\\
\rightarrow a=\pm 1\,\, or \,\, a=\pm 2--------(1)\\
(a-y)[a(b-2)-\dfrac{28}{y}]=0---(C)\\
\therefore a=y=\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28---(2)$
for $a,b\in Z$
check $(1)(2)$
the only solution :$a=2,b=-5$ will satisfy (A)
by symmetry:
$\therefore (a,b)=(2,-5) \,\, or \,\, (a,b)=(-5,2)$

 

[anemone]

Thanks Albert for participating, and your final solution is correct.

My solution:

Spoiler

$a^2(b-2)+b^2(a-2)+28=0$

$a^2(b-2)+b^2(a-2)+32=4$

$a^2b-2a^2+ab^2-2b^2+32=4$

$ab(a+b)-2(a^2+b^2)+32=4$

$ab(a+b)-2(a+b)^2+4ab+32=4$

$(a+b)(ab-2(a+b))+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)-8(a+b)+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)+4(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=\pm 1(\pm 1)\stackrel{\text{or}}=\pm 2(\pm 2)\stackrel{\text{or}}=\pm 4(\pm 1)\stackrel{\text{or}}=\pm 1(\pm 4)$

Upon checking for all cases, we see that the following two are the only solutions:

$(a,\,b)=(-5,\,2),\,(2,\,-5)$

 

[kaliprasad]

Thanks Albert for participating, and your final solution is correct.

My solution:

Spoiler

$a^2(b-2)+b^2(a-2)+28=0$

$a^2(b-2)+b^2(a-2)+32=4$

$a^2b-2a^2+ab^2-2b^2+32=4$

$ab(a+b)-2(a^2+b^2)+32=4$

$ab(a+b)-2(a+b)^2+4ab+32=4$

$(a+b)(ab-2(a+b))+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)-8(a+b)+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)+4(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=1(1)\stackrel{\text{or}}=2(2)\stackrel{\text{or}}=4(1)\stackrel{\text{or}}=1(4)$

Upon checking for all cases, we see that the following two are the only solutions:

$(a,\,b)=(-5,\,2),\,(2,\,-5)$

above ans is correct nut

Spoiler

you missed out some cases of factors (-ve) * (-ve) in which you could have missed out solution but you did not miss out

 

[anemone]

Ah, my mistake...I in fact took into consideration of all cases, just that I forgot to put the plus minus sign in the post...I just fixed it, thanks!

 

[kaliprasad]

Thanks Albert for participating, and your final solution is correct.

My solution:

Spoiler

$a^2(b-2)+b^2(a-2)+28=0$

$a^2(b-2)+b^2(a-2)+32=4$

$a^2b-2a^2+ab^2-2b^2+32=4$

$ab(a+b)-2(a^2+b^2)+32=4$

$ab(a+b)-2(a+b)^2+4ab+32=4$

$(a+b)(ab-2(a+b))+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)-8(a+b)+4(ab+8)=4$

$(a+b)(ab-2(a+b)+8)+4(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=4$

$(a+b+4)(ab-2(a+b)+8)=\pm 1(\pm 1)\stackrel{\text{or}}=\pm 2(\pm 2)\stackrel{\text{or}}=\pm 4(\pm 1)\stackrel{\text{or}}=\pm 1(\pm 4)$

Upon checking for all cases, we see that the following two are the only solutions:

$(a,\,b)=(-5,\,2),\,(2,\,-5)$

Sorry that I missed out last time

Spoiler

$\pm1(\pm 1)$ is invalid