(a+b)^k-a^k-b^k

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Source: PRMO 2019 Leg 2 P8

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#1 h Aug 25, 2019, 5:16 PM • 3 Y

Y by Physimathematics, Adventure10, Mango247

Let $F_k(a,b)=(a+b)^k-a^k-b^k$ and let $S={1,2,3,4,5,6,7,8,9,10}$ . For how many ordered pairs $(a,b)$ with $a,b\in S$ and $a\leq b$ is $\frac{F_5(a,b)}{F_3(a,b)}$ an integer?

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#2 h Aug 25, 2019, 5:59 PM • 3 Y

Y by Pluto1708, Physimathematics, Adventure10

Since it is impossible that $a=-b$ , we find that
$\begin{align*} \frac{F_5 (a,b)}{F_3 (a,b)} &= \frac{(a+b)^5-a^5 - b^5}{(a+b)^3-a^3 - b^3}\\ &=\frac{5ab(a+b)(a^2 + ab + b^2)}{3ab(a+b)}\\ &= \frac{5(a^2 + ab + b^2)}{3}. \end{align*}$
So we want $a^2 + ab + b^2 \equiv 0 \pmod{3}.$ Multiplying both sides by $a-b$ yields $a^3 \equiv b^3 \pmod {3}$ . We find the classes
$\begin{align*} 1^3 \equiv 4^3 \equiv 7^3\equiv 10^3 \equiv 1 &\pmod{3}\\ 2^3 \equiv 5^3 \equiv 8^3 \equiv -1 &\pmod{3}\\ 3^3 \equiv 6^3 \equiv 9^3 \equiv 0 &\pmod{3}. \end{align*}$
We suspect that the answer is $4^2 + 3^2 + 3^2= \boxed{34}$ and indeed we check that picking any two numbers in a class works, as any two numbers in a class is separated by a multiple of $3$ :
$(a+3n)^2 + (a+3n)a + a^2 = 3(a^2 + 3an + n^2)\equiv 0\pmod{3}.$
EDIT: I didn't see the $a\leq b$ condition. We actually have $\binom{4}{2}+\binom{3}{2}+\binom{3}{2}=12$ choices for $a<b$ by choosing pairs that live in the same class and then add $10$ of the $a=b$ to get $\boxed{22}$ .

This post has been edited 1 time. Last edited by BigSams, Aug 25, 2019, 6:17 PM

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#3 h Aug 25, 2019, 6:01 PM • 2 Y

Y by Physimathematics, Adventure10

Could you please write out all the pairs? Because I am getting 22.
Maybe you missed $a\leq b$ . I don't know. I can't understand anything from your solution sir

BTW plz accept friend request

This post has been edited 3 times. Last edited by Math-wiz, Aug 25, 2019, 6:02 PM

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#4 h Aug 25, 2019, 6:17 PM • 2 Y

Y by Physimathematics, Adventure10

Thanks, I did forget the $a\le b$ condition. Your answer is right and I have edited my post.

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#5 h Aug 27, 2019, 9:29 AM • 3 Y

Y by Physimathematics, Adventure10, Mango247

Very short solution , thx @BigSams , it can be made more shorter just saying that a<b and b<a cases are symmetric , so for each there are (34-10)/2 = 12 cases, hence 22 answer

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#6 h Feb 26, 2020, 7:22 AM • 1 Y

Y by Physimathematics

If (a,b) both are of form either $3k$ or $3k+1$ the $a^2+b^2+ab$ is divisible by 3.

Now $3k\in \{3,6,9\}$ and $3k+1\in \{1,4,7,10\}$ and note $a,b=(2,2),(5,5),(8,8)$ are also solution . so number of order pair is $3^2+4^2+3=28$ .

This post has been edited 2 times. Last edited by ftheftics, Mar 2, 2020, 6:30 AM
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#8 h Feb 26, 2020, 7:29 AM • 2 Y

Y by Physimathematics, Mango247

ftheftics wrote:

If (a,b) both are of form either $3k$ or $3k+1$ the $a^2+b^2+c^2$ is divisible by 3.

Now $3k\in \{3,6,9\}$ and $3k+1\in \{1,4,7,10\}$ and note $a,b=(2,2),(5,5),(8,8)$ are also solution . so number of order pair is $3^2+4^2+3=28$ .

Note that $a\le b$ .

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#9 h Feb 28, 2020, 1:28 PM • 2 Y

Y by Physimathematics, Mango247

Ooh I got it .

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MATHS_ENTUSIAST

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#10 h Apr 21, 2025, 1:26 PM

Y by

I just simplified and got
ab|5(a²+b²)
And I did caseworking on a=1,2,3,....,10
And got 22 pairs

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