# What will be the value of $ab$?

I have:

If $$a$$ and $$b$$ are positive integers such that $$a^3-b^3=61$$, then the value of $$ab$$ is?

(1) $$20$$

(2) $$15$$

(3) $$35$$

(4) $$63$$

Now, I know the identity $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ so $$61 = (a - b)(a^2 + ab + b^2).$$ But what to do next? I have a solution in which it is written, " Here , $$(a - b) = 1$$ [ because $$a$$ and $$b$$ are positive integers ] But how can the difference of any two unknown integers be $$1$$?

• Hint: mod $2$ it's $\,a-b \equiv 1$ so $a,b$ have opposite parity so their product is even, so it's $(1)\$ – Bill Dubuque Oct 26 '18 at 19:40

$$61=(a-b)(a^2+ab+b^2)$$, Since $$a^2+ab+b^2>0$$ you know that $$a>b$$ otherwise $$(a-b)<0$$.

Next since $$a$$ and $$b$$ are positive integers $$a^2+ab+b^2$$ is a positive integer call it $$c$$. Now you need to realize that $$\frac{61}{a-b}=c$$.

However 61 is a prime number. Thus $$a-b=1$$.

The choices B,C,D do not have factors whose difference is 1. For example $$15=3\times 5,1\times 15$$. $$5-3=2$$ and $$15-1=15$$. Thus B cannot be the answer. similar for the other ones.

• Thank you so much :) – yena shah Oct 26 '18 at 4:07
• @yenashah Much simpler to use a parity argument to show that $a$ or $b$ is even hence so is their product - see my answer for a generalization. – Bill Dubuque Oct 26 '18 at 20:05

Notice that $$61$$ is a prime number.

Hence we have $$a-b=1$$ and $$a^2+ab+b^2=61$$.

$$a=b+1$$. $$(b+1)^2+(b+1)b+b^2 = 61$$

$$3b^2+3b-60=0$$

$$b^2+b-20=0$$

Hence you should be able to solve for $$b$$ and recover $$a$$ as well.

Another "quick" way which is good technique for multi-choice questions.

After you realise that $$a-b=1$$ based on the primality of $$61$$, you get $$a^2 + ab + b^2 = 61$$. You can rearrange the left hand side to $$(a+b)^2 - ab$$, so you get $$61 + ab = (a+b)^2$$. You're now asking yourself - which number added to $$61$$ gives you a perfect square? The almost immediate answer is - only the first option ($$61+20 = 81 = 9^2$$).

$$a,b$$ aren't both odd, else $$\,a^3\!-b^3\,$$ is even $$\neq 61$$. So at least one of $$a,b$$ is even so $$ab$$ is even, so it's ($$2)$$

More generally if $$\,f(x,y)\,$$ is a polynomial with integer coefficients and $$\,f(1,1)\,$$ is odd then $$f(a,b)=0\,\Rightarrow\,2\mid ab,\,$$ else $$\,a,b\,$$ are odd so $$\!\bmod 2\!:\ a,b\equiv 1\,\Rightarrow\, 0 = f(a,b)\equiv f(1,1)\equiv 1$$

Here $$\,f(x,y) = x^3-y^3-61\,$$ so $$\,f(1,1) = -61\,$$ is odd, so above is a special case.

This is a bivariate form of the polynomial Parity Root Test.

61 is a prime number, which can be factorized in a unique way: $$61*1$$. Your 2 expressions in () are integers since a and b are integers, hence you can only have 2 scenarios:

1. first () = 61 and second () = 1 (impossible since $$a^2 + ab + b^2 > 1$$

2. first () = 1 and second () = 61.

Here's one way to do it:

$$a-b$$ has to be odd, because if $$a-b$$ were even, $$(a-b)(a^2+b^2+ab)$$ would be even. This eliminates (2),(3),and (4) because 15, 35, and 63 only have odd factors, so the difference of those factors will be even.

15: 1,3,5,15

35: 1,7,5,35

63: 1,3,9,7,21,63

• This is a special case of the Parity Root Test - see my answer. – Bill Dubuque Oct 26 '18 at 20:02

Here, a^3-b^3= (a-b)(a^2+b^2+ab)=61 As 61 is prime clearly a-b=1 Now, a^2+b^2+ab=61 (a+b)^2-ab=61 (2b+1)^2-ab=61 (2b+1)^2-b(1+b)=61 Solving b=4 and hence a=5 Hence ab=20