$x^2-y^2=196$, can we find the value of $x^2+y^2$?

Question

$x$ and $y$ are positive integers.

If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?

Can anyone explain this to me? Thanks in advance.

Answer 1

Rewrite the equation to get $x^2=y^2+196$. Now write $x=y+k$, then $$x^2=(y+k)^2=y^2+2yk+k^2=y^2+196$$

Therefore $2yk+k^2=196$

Now note that $k \mid 196$ and $k$ must be even.

This leaves the possibilities $2,4,14,28,98,196$.

$k=2$ gives $y=48$.

$k=4$ gives $y=22.5$, which is not an integer.

$k=14$ gives $y=0$, but that is not positive.

Similiar, for larger $k$, it won't be positive.

So $y=48$, therefore $$x^2+y^2=(x^2-y^2)+2y^2=196+2\cdot48^2=4804$$

Answer 2

Hint 1: $x^2-y^2 = (x+y)(x-y)$.

Hint 2: Divisors of $196$: $1$, $2$, $4$, $7$, $14$, $28$, $49$, $98$, $196$.

Answer 3

It is possible to find out $x^2+y^2$, as $196$, which will be $(x-y)(x+y)$, can be factorised as five possible integer products, $14\times 14$, $7\times28$, $4\times 49$, $2\times 98$, $1\times 196$.

Of these only one product leads to both positive and integer solutions for $x$ and $y$. Can you work out which of the products this will be?

Answer 4

Just use Pythagora's theorem:$$x^2=y^2+(14)^2$$

$$y=p^2-q^2$$ $$z=2pq=14$$ $$x=p^2+q^2$$ Since $2pq=14$ and $p,q$ are coprime, we can only have $(p,q)=(\pm1,\pm7);(\pm7,\pm 1)$ From here, it is easy: $$x^2+y^2=50^2+48^2= 4804 $$

Answer 5

$$ x^2-y^2=(x-y)(x+y)=14^2 $$ Since $(x+y)-(x-y)=2y$ either both $x+y$ and $x-y$ need to be even or both need to be odd. Since their product is even, both need to be even. There are two possible factorizations: $14\cdot14$ and $2\cdot98$. $14\cdot14$ implies $y=0$, so we want $x-y=2$ and $x+y=98$. That is, $x=50$ and $y=48$. Therefore, $$ x^2+y^2=4804 $$