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

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?

• so $(a+b)(a-b)=37$, and $37$ is prime, so $a+b=?$, $a-b=?$, $a=?$, $b=?$, $a^2+b^2=?$ Commented Mar 18 at 12:31
• That is a remarkable guess. Commented Mar 18 at 12:31
• Prime factorization; b can't be zero Commented Mar 18 at 12:33
• Indeed: $361+324=685$ :-) Commented Mar 18 at 12:37
• I don't think there is any other way than noting that $a^2-b^2$ factors, 37 is prime and figuring the only choices of $a,b$. I feel bad about these questions as the require either insight of familiarity with similar and neither can be taught. Commented Mar 19 at 15:24

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.

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

• Why isn't $\{ |a-b|, |a+b|\} = \{37 , 1\}$ valid? Commented Mar 19 at 18:21
• @GameTimeWithAryan It is valid. Note that when you represent a set by listing all its elements, the order does not matter. Particularly, $\{1,37\} = \{37,1\}$. Maybe you are confusing it with the $\Bbb R^2$ vector $(1,37)$.
– D S
Commented Mar 19 at 18:25
• I think case work can be avoided also by stating "WOLOG we may assume $a,b\in \mathbb N$". Of course, one of my peeves are folks who state wolog when not making it clear why we can wolog. In this case as our only concern is with the squares of $a$ and $b$ we can simply assume we use that positive values. (We could just as well have $a=-19$ and $b=+18$) Commented Mar 20 at 15:23

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, $$a^2 + b^2 = 361 + 324 = 685$$

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.

• I think from "$a^2-b^2=(b+d)^2-b^2=2bd+d^2=37$" also has the simplification at the right of $d*(2b+d)=37$. Since $a^2+b^2$ can be evaluated with Case 1: $a>0, b>0$ or Case 2 $a<0, b<0$, Each Case yielding the same answer for $a^2+b^2$. Hence, start with all natural numbers $d>0$ and $2b+d>0$, as possible natural prime factors of $37$. Clearly with $(2b+d)$ as a natural number $(2b+d)>d$. With $(2b+d)$ as a greater natural number factor and $d$ as a lesser natural number factor, it follows that $d=1$. And also $2b+1=37 \implies b=18$ and $a=b+d=19$. Then $a^2+b^2=18^2+19^2=685$! Commented Mar 20 at 8:09
• I did not yet create an edit with the previous comment, but I would appreciate if someone like the author or someone else edits the original posts the the above shortcut, since comments can be easily deleted and posts not-so. I do not want to offend the original contributor who helped so much so I have decided not to edit his work. Now with the comment, I am giving this answer also an up-vote since it is an original perspective towards the question and answer. Commented Mar 20 at 8:13
• @StephenElliott I don't understand the purpose of either of you comments. I think it compounds things to consider insignificant cases when they can be explained as insignificant with a "wolog" expression (hence my parenthetical. And knowing that $d|37=2bd+d^2$ doesn't require expressing the other factor, $2b+d$ and evaluating it as larger than $d$ when we know $d \le 6 < 37$. From that alone it follows $d=1$. Commented Mar 20 at 15:14

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