Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.
Prove that $a^2+b^4=1994$.
[sp]If $a^2+b^4=2\bigl((a-6)^2+(b+1)^2\bigr) = 2a^2 + 2b^2 - 24a + 4b + 74$ then $b^4 - 2b^2 - a^2 + 24a = 4b+74.$ Complete the squares on the left, to get $(b^2-1)^2 - (a-12)^2 = 4b-69.$
The two squares on the left cannot be equal, because that would mean $4b-69=0$, and that does not have an integer solution. So there are two possible cases.
Case 1: $b^2-1 > a-12$. In this case, $a-12$ is at most $b^2-2$. Therefore $4b-69 = (b^2-1)^2 - (a-12)^2 \geqslant (b^2-1)^2 - (b^2-2)^2 = 2b^2-3.$ But this says that $0\geqslant 2b^2 - 4b + 66 = 2(b-1)^2 + 64.$ That is clearly impossible, so this case cannot arise.
Case 2: $b^2-1 < a-12$. Then $a-12$ is at least $b^2$, and so $4b-69 =(b^2-1)^2 - (a-12)^2 \leqslant (b^2-1)^2 - (b^2)^2 = 1-2b^2.$ Therefore $2b^2 + 4b - 70 \leqslant0.$ This says that $2(b+1)^2 \leqslant 72$, so that $(b+1)^2 \leqslant 36$, $b+1 \leqslant 6$, $b\leqslant5.$ But we are told that $b>3$. Therefore $b=4$ or $5$.
If $b=4$ then the equation $(b^2-1)^2 - (a-12)^2 = 4b-69$ becomes $(a-12)^2 = 278.$ But that is not a perfect square, so we must have $b=5$, in which case $(a-12)^2 = 625 = 25^2$, and $a = 25+12 = 37.$
To prove:
$a^2+b^4=2((a-6)^2+(b+1)^2)=1994$.
that is to prove:
$(a-6)^2+(b+1)^2=997=31^2+6^2$
$\therefore a=37,\,\, b=5$
[sp]This verifies that $a=37$, $b=5$ is a possible solution of the equation $a^2+b^4=2\bigl((a-6)^2+(b+1)^2\bigr)$. But it does not show that it is the only one. It assumes that both sides are equal to 1994, which is the result that is to be proved.[/sp]
To prove:
$a^2+b^4=2((a-6)^2+(b+1)^2)=1994$.
that is to prove:
$(a-6)^2+(b+1)^2=997=31^2+6^2$
$\therefore a=37,\,\, b=5$
Hey Albert, like Opalg has already mentioned, you have to prove that $(a,\,b)=(37,\,5)$ is the only solution to that original given equation.:)
Opalg said:
[sp]If $a^2+b^4=2\bigl((a-6)^2+(b+1)^2\bigr) = 2a^2 + 2b^2 - 24a + 4b + 74$ then $b^4 - 2b^2 - a^2 + 24a = 4b+74.$ Complete the squares on the left, to get $(b^2-1)^2 - (a-12)^2 = 4b-69.$
The two squares on the left cannot be equal, because that would mean $4b-69=0$, and that does not have an integer solution. So there are two possible cases.
Case 1: $b^2-1 > a-12$. In this case, $a-12$ is at most $b^2-2$. Therefore $4b-69 = (b^2-1)^2 - (a-12)^2 \geqslant (b^2-1)^2 - (b^2-2)^2 = 2b^2-3.$ But this says that $0\geqslant 2b^2 - 4b + 66 = 2(b-1)^2 + 64.$ That is clearly impossible, so this case cannot arise.
Case 2: $b^2-1 < a-12$. Then $a-12$ is at least $b^2$, and so $4b-69 =(b^2-1)^2 - (a-12)^2 \leqslant (b^2-1)^2 - (b^2)^2 = 1-2b^2.$ Therefore $2b^2 + 4b - 70 \leqslant0.$ This says that $2(b+1)^2 \leqslant 72$, so that $(b+1)^2 \leqslant 36$, $b+1 \leqslant 6$, $b\leqslant5.$ But we are told that $b>3$. Therefore $b=4$ or $5$.
If $b=4$ then the equation $(b^2-1)^2 - (a-12)^2 = 4b-69$ becomes $(a-12)^2 = 278.$ But that is not a perfect square, so we must have $b=5$, in which case $(a-12)^2 = 625 = 25^2$, and $a = 25+12 = 37.$
I will show another method (which isn't my solution) here, so that we know this problem admits at least two methods to solve it.
Rewriting the given equation we have, a quadratic in $a$ where
$a^2-24a-b^4+2b^2+4b+74=0$
It has integer solutions only if the discriminant $4(b^4-2b^2-4b+70)$ is a perfect square. It's easy to show that for $b\ge 4$,
$(b^2-2)^2<b^4-2b^2-4b+7<(b^2+1)^2$
Indeed, $b^2-2b+33>0$ and $4b(b+1)>69$. The first inequality is true. Since $b\ge 4$, $4b(b+1)>4\cdot 4\cdot 5=80>69$. The only perfect perfect squares between $(b^2-2)^2$ and $(b^2-1)^2$ are $(b^2-1)^2$ and $(b^2)^2$. Now,
$(b^2-1)^2=b^4-2b^2-4b+70$ gives $b=\dfrac{69}{4}$ which isn't an integer.
$b^4-2b^2-4b+70=b^4$ gives $b^2+2b-35=0$ or $b=5$ or $b=-7$.
Thus, $b=5$ and it gives $a=37$ and $a^2+b^4=37^2+5^4=1994$.
