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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) $63$

( Answer : option(1) )

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 $1$?

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    $\begingroup$ Hint: mod $2$ it's $\,a-b \equiv 1$ so $a,b$ have opposite parity so their product is even, so it's $(1)\ $ $\endgroup$ – Bill Dubuque Oct 26 '18 at 19:40
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$61=(a-b)(a^2+ab+b^2)$, 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.

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  • $\begingroup$ Thank you so much :) $\endgroup$ – yena shah Oct 26 '18 at 4:07
  • $\begingroup$ @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. $\endgroup$ – Bill Dubuque Oct 26 '18 at 20:05
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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.

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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 ($61+20 = 81 = 9^2$).

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$a,b$ 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.

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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.

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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

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

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