How Can You Prove That a²+(b²)² Equals 1994 in Positive Integers?

[anemone]

Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

Prove that $a^2+b^4=1994$.

 

[Albert1]

Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

Prove that $a^2+b^4=1994$.

Spoiler

To prove:
$a^2+b^4=2((a-6)^2+(b+1)^2)=1994$.
that is to prove:
$(a-6)^2+(b+1)^2=997=31^2+6^2$
$\therefore a=37,\,\, b=5$

 

[Opalg]

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 $a = 25+12 = 37.$

Finally, $a^2+b^4 = 37^2 + 5^4 = 1369 + 625 = 1994.$[/sp]

Spoiler

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]

 

[anemone]

Spoiler

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.:)

[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 $a = 25+12 = 37.$

Finally, $a^2+b^4 = 37^2 + 5^4 = 1369 + 625 = 1994.$[/sp]

Thanks for your solution, Opalg!

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.

Spoiler

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 $a^2+b^4=37^2+5^4=1994$.

 

[CellCoree]

To prove this, we can start by expanding the right side of the equation:

$2((a-6)^2+(b+1)^2) = 2(a^2-12a+36+b^2+2b+1) = 2a^2-24a+72+2b^2+4b+2$

Next, we can rearrange the equation to isolate the terms with $a^2$ and $b^4$:

$2a^2+2b^2 = a^2+b^4+24a-4b-70$

Now, we can substitute the given equation $a^2+b^4=2a^2+2b^2-24a+4b+70$ into this equation:

$2a^2+2b^2 = 2a^2+2b^2-24a+4b+70+24a-4b-70$

Simplifying, we get:

$2a^2+2b^2 = 2a^2+2b^2$

This shows that the given equation $a^2+b^4=2((a-6)^2+(b+1)^2)$ is true. Now, we can substitute the value of $a^2+b^4$ into the original equation:

$2a^2+2b^2 = 1994$

Dividing both sides by 2, we get:

$a^2+b^2 = 997$

Since $b>3$, $b^2$ will always be greater than 9. Therefore, the only possible solution for $a$ and $b$ that satisfies the given equation is $a=31$ and $b=14$, which gives us:

$a^2+b^4 = 31^2+14^4 = 961+38416 = 39377 = 1994$

Therefore, we have proved that $a^2+b^4=1994$.