$a^2-b^2=37$, evaluate $ a^2+b^2$ [closed]

Question

Given $a^2-b^2=37$ and also a and b are integers, can we evaluate $a^2+b^2$ possible values? Are those many or just some? I found that $a^2+b^2$ can be only $685$. But how to prove it? I just guessed, but can we somehow evaluate it?

Answer 1

The product of $a+b$ and $a-b$ is $37$, and since $37$ is prime,

we must have one factor $1$ and the other $37,$ or one factor $-1$ and the other $-37$.

Now solve for $a$ and $b$ in each of these cases and see what $a^2+b^2$ is.

Answer 2

Casework can be avoided as follows:

It has already been noted that 37 is a prime, and $(a+b)(a-b) = 37$ for integers $a,b$ implies that $\{ |a-b|, |a+b|\} = \{1 , 37\}$.

Then $$a^2+b^2 = \frac{|a-b|^2+|a+b|^2}{2} = \frac{1^2+37^2}{2} = \boxed{685}$$

Answer 3

I am just building upon J.W. Tanner's comment and posting it as an answer.

Your equation can be rewritten as $(a+b)(a-b)=37$ and since 37 is prime we have four cases.

Case $1$ : $a+b=37$ and $a-b=1$ $\Rightarrow a = 19,\ b=18$
Case $2$ : $a+b=1$ and $a-b=37$ $\Rightarrow a = 19,\ b=-18$
Case $3$ : $a+b=-37$ and $a-b=-1$ $\Rightarrow a = -19,\ b=-18$
Case $4$ : $a+b=-1$ and $a-b=-37$ $\Rightarrow a = -19,\ b=18$

In all cases, $a^2 + b^2 = 361 + 324 = 685$

Answer 4

I present an alternative. It is not as good an alternative. In fact I would say this is a bad way to solve. But I want to impress that if one doesn't always see the correct insight one can still chew an answer and there is always more than one way to do things.

$a^2 -b^2 =37$ so $a^2 > b^2$ and if we assume $a,b$ are positive integers (we might as well, whether they are negative of positive their squares are positive and we are only answering questions about their squares) then $a > b$. Let $d = a-b$ so $a=b+d$.

Then we have $a^2 -b^2 = (b+d)^2 - b^2 = 2bd + d^2 = 37$.

We see that $d^2 < 2bd + d^2 =37$ so $d= 6,5,4,3,2,1$ and we can test each value. If that is too much work, we can see that $2bd$ is even and $37$ is odd so $d$ is odd and we can test $d = 5,3, 1$. If that is still too much work we can not that $d|d^2$ and also $2bd$ so $d|2bd +d^2 = 37$. But $37$ is prime so $d=1$.

And so $2bd + d^2 = 2b + 1=37$ and $b=18$. ANd $a = b+d = 18+1=19$.

Which.... if we were really paying attention we might have noticed we were skirting the other better way of doing it. We noted that $d|2bd +d^2$ but as $d =b-a$ and $2bd+d^2 = a^2 -b^2$ we are noting that $a-b|a^2-b^2$ and as $\frac {2bd + d^2}d = 2b + d = b + (b+d) = b + a = a +b$ we were really just doing the same thing as everyone else.

Answer 5

To restate the question for clarity:

$$ \boxed{ \text{Given }a^2-b^2=37\text{ and }a\text{ and }b\text{ are integers; Then, evaluate }a^2+b^2 .}\tag{Eq. 1}$$ Add $b^2$ to both the left and the right of Equation 1, and solve for $a^2$ in terms of $b^2$, and then sum $a^2$ and $b^2$ so: $$ a^2=b^2+37 \underset{implies}\implies a^2+b^2=(b^2+37)+b^2 = 2*b^2+37 \tag{Eqs. 2} $$ Alternatively, given a value of $a^2$ it is possible to determine also $a^2+b^2$ in terms of $a^2$ as follows. Subtract from the left of Equations 2 the value $37$, so that: $$ b^2=a^2-37 \underset{implies}\implies a^2+b^2=a^2+(a^2-37)=2*a^2-37 \tag{Eqs. 3} $$ So given any value for $a^2$ or $b^2$, it is possible to find the value for $a^2+b^2$ (from Equations 2 and Equations 3) as: $$ \text{Given }a\text{ then }a^2+b^2=2*a^2-37 \text{ or } \text{Given }b\text{ then }a^2+b^2=2*b^2+37 \tag{Eqs. 4} $$ Now comes the restriction that $a$ and $b$ are integers. Given $a$,Equations 4 on the left imply: $$ (b-a)(b+a)=-37 \underset{implies}\implies (a-b)(a+b)=37 \tag{Eq. 5} $$ But $37$ is prime!, meaning it can be factored by $1$ and itself $37$, according to the definition of a prime number:

That would lose the useful idea of a prime number. We could also say that 1/2 is a factor of 5. So we restrict the possible factors to positive integers.

The question has that $a$ and $b$ are integers and thus can be positive or negative. However, if $a>0$ then also $b>0$. And if $a<0$ then also $b<0$ so as to preserve the property that $((a+b)*(a-b)=37)>0$.

In either case, the factor $(a+b)$ can be assigned a value, and then it follows what the value for $(a-b)$ is. Case 1: For instance, when $(a+b)>0$ set $(a+b)=37$, since here $(a+b)>(a-b)$. So that $(a+b)*(a-b)>0$ and $((a+b)>0) \underset{implies}\implies (a-b)>0$ also, set $(a-b)=1$.

Alternatively, there is this Case 2: $(a+b)<0$ and $(a-b)<0$ so that $((a+b)*(a-b)>0)=37$, when $(a+b)<0$ set $(a+b)=-37$, since here $(a+b)<(a-b)$. So that $(a+b)*(a-b)>0$, also set $(a-b)=-1$.

Now we seek an identity for $a^2+b^2$ in terms of $a+b$ and $a-b$. Consider:

$$\frac{(a+b)^2+(a-b)^2}{2}=\frac{(a^2+2*a*b+b^2)+(a^2-2*a*b+b^2)}{2}=\frac{(a^2+b^2+2*a*b+b^2)}{2}+\frac{(a^2+b^2-2*a*b+b^2)}{2} =a^2+b^2 \tag{Eqs. 6}$$ This is the identity needed, namely $a^2+b^2=\frac{(a+b)^2+(a-b)^2}{2}$ and this identity does not depend on the sign of $(a+b)$ or of $(a-b)$ even since both quantities are squared!

So, finally, $$ \boxed{ a^2+b^2=\frac{37*37+1*1}{2}=\frac{1369+1*1}{2}==\frac{1370}{2}=685 } \tag{Eqs. 7} $$