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The following expression yields an integer answer (very surprisingly it seems).

$${\left (515391\frac{33}{40} \right)}^4 - {\left (384140\frac{39}{40} \right)}^4 = 48783404650404592720562$$

I have tried many other pairs of 6-digit integers , but none of them result in an integer answer. For example: $${\left (515390\frac{33}{40} \right)}^4 - {\left (384141\frac{39}{40} \right)}^4 = 48782630297643606282783.184$$

$${\left (515389\frac{33}{40} \right)}^4 - {\left (384142\frac{39}{40} \right)}^4 = 48781855946299371591451.728$$

$${\left (515392\frac{33}{40} \right)}^4 - {\left (384139\frac{39}{40} \right)}^4 = 48784179004582352493575.376$$

I have tried over a hundred pairs of 6-digit integers , but none of them result in an integer answer. It seems that a special property or characteristic of the pair of integers 515391 and 384140 , makes the above expression an integer. But what special property or characteristic ? Can anyone see why the above expression is an integer ?

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closed as unclear what you're asking by Namaste, Jack, Daniel W. Farlow, I am Back, user223391 Jul 1 '17 at 22:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Tip for future enclose math with dollar signs \$\$ otherwise it looks like a bunch of gibberish. $\endgroup$ – kingW3 Jun 26 '17 at 23:03
  • $\begingroup$ Do you mean $\dfrac{(33)^4}{40}$ or $\left(\dfrac{33}{40}\right)^4$ The edit favors the first interpretation. Just wanted to check with you, too, Derek. $\endgroup$ – Namaste Jun 26 '17 at 23:06
  • $\begingroup$ Have you made a calculation mistake? For @amWhy 's first and second expressions, none of them (wolframalpha.com/input/…), (wolframalpha.com/input/…) give the answer that you have. $\endgroup$ – Toby Mak Jun 26 '17 at 23:08
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    $\begingroup$ They're mixed fractions, with the exponent applying to the entire mixed fraction. I'll edit the question to reflect this. $\endgroup$ – Tanner Swett Jun 26 '17 at 23:09
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    $\begingroup$ There are many others, but they are not all that common: e.g. $\left(151359\frac{39}{40}\right)^4-\left(106406\frac{17}{40}\right)^4=396665766866791017179$ $\endgroup$ – robjohn Jun 27 '17 at 15:56
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Let $x=40(515391\frac{33}{40})$ and $y=40(384140\frac{39}{40})$. Your expression will be an integer if and only if $x^4-y^4$ is a multiple of $40^4=2^{12}\cdot 5^4$.

The fact that it is a multiple of $40^4$ can be seen by observing the following three things:

  • Both $x$ and $y$ are odd.
  • Their sum $x+y$ is a multiple of $2^{10}$. (Actually, it's a multiple of $2^{11}$, which is why the integer you got is even, but $2^{10}$ would be good enough.)
  • There is a number $q$ (in this case $q=2$) such that $x \equiv 7q \pmod{5^4}$ and $y \equiv 24q \pmod{5^4}$.

Why is this enough? It boils down to the factorization $x^4-y^4=(x-y)(x+y)(x^2+y^2)$:

  • Because $x$ and $y$ are both odd, $x-y$ and $x^2+y^2$ are both even. As $x+y$ is a multiple of $2^{10}$, $(x-y)(x+y)(x^2+y^2)$ is a multiple of $2^{12}$.
  • Because $x \equiv 7q \pmod{5^4}$ and $y \equiv 24q \pmod{5^4}$, and $(7,24,25)$ is a pythagorean triple, we have: $$x^2+y^2\equiv (7^2+24^2)q^2 =25^2q^2\equiv0 \pmod{5^4}$$

So $x^4-y^4$ is a multiple of both $2^{12}$ and $5^4$, which is what we wanted.


Note that these three conditions are general enough to let you find lots of other pairs $(x,y)$ with $x^4-y^4$ a multiple of $40^4$ (though not all of them):

  • Start by choosing $N=x+y$, a multiple of $2^{10}$. For best results, it should be coprime to $5$ (otherwise, the fractions you get won't have 40 in the denominator in reduced form, so it won't look as impressive).
  • Since $31$ is coprime to $5$, the equation $N \equiv (7+24)q=31q \pmod{5^4}$ can be solved for $q$. Take $q$ to be a solution to that equation, and choose some odd $x \equiv 7q \pmod{5^4}$.
  • Finally, set $y=N-x$. Then $x$ and $y$ satisfy the top three bullet points in this answer, so $x^4-y^4$ will be a multiple of $40^4$.

For example, we could take $N=2^{10} \cdot 123456789 = 126419751936$. Then $N \equiv 61 \pmod{5^4}$, and so we have to solve the equation $61 \equiv 31q \pmod{5^4}$. This equation has solution $q=506$, so we can take any odd $x \equiv 7 \cdot 506 \equiv 417 \pmod{5^4}$. Let's take $x=417 + 5^4 \cdot 98765432 = 61728395417$. Finally, $y=N-x=64691356519$. You can check that $$ \left(\frac{x}{40}\right)^4-\left(\frac{y}{40}\right)^4=\left(1543209885\frac{17}{40}\right)^4-\left(1617283912\frac{39}{40}\right)^4 $$ is an integer.


Based on your comment, you are specifically interested in $x,y$ which are congruent to 33 and 39 modulo 40. You can do that in the same way, but it takes a little extra work. Since $N=x+y$, we need to take $N \equiv 33+39 \equiv 32 \pmod{40}$ as well as being a multiple of $2^{10}$. It's not hard to check that this happens when $N=2^{10} \cdot p$ where $p \equiv 3 \pmod{5}$. For example, we could take $N=2^{10} \cdot 1000003=1024003072$.

You can check that, if we take $q=462$, then $31q \equiv N \pmod{5^4}$. So we want to choose $x$ with $x \equiv 7 \cdot 462 \equiv 109 \pmod{5^4}$ and either $x \equiv 33$ or $x \equiv 39 \pmod{40}$. The first of these is impossible (because it implies incompatible things about $x$'s mod 5 congruence class) but the second is possible: it happens whenever $x \equiv 1359 \pmod{5000}$. For example, we could take $x=400001359$, and then $y=1024003072 - x = 624001713$. Again, you can check that $$ \left(\frac{x}{40}\right)^4 - \left(\frac{y}{40}\right)^4 = \left(10000033 \frac{39}{40}\right)^4- \left(15600042\frac{33}{40}\right)^4\\ = -49224604028046128106617226275 $$ is an integer, which after swapping the order of the terms gives you what you want.

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  • $\begingroup$ Dear @Micah . All of what you say is correct. But I am looking for expressions of the form (an integer frac{33}{40})^4 - (an integer frac{39}{40})^4= an integer . Your example gives us a way of finding an expression of the form (an integer frac{17}{40})^4 - (an integer frac{39}{40})^4= an integer . This is not what I am looking for !! $\endgroup$ – Derek Jun 27 '17 at 7:27
  • $\begingroup$ @Derek: See my edit. $\endgroup$ – Micah Jun 27 '17 at 18:13
  • $\begingroup$ Excellent piece of work. Great Help. This is just what I was looking for !! Thanks very much @Micah!! $\endgroup$ – Derek Jun 27 '17 at 18:59
  • $\begingroup$ @Derek: If you like Micah's answer, kindly click on the button to accept it. $\endgroup$ – Tito Piezas III Jun 28 '17 at 2:43
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I don't have a great answer, but here's what I've been able to come up so far.

You've found that

$${\left (515391\frac{33}{40} \right)}^4 - {\left (384140\frac{39}{40} \right)}^4$$

is an integer. This is equivalent to saying that the above number, times $40^4$, is a multiple of $40^4$. In other words,

$${(40 \cdot 515391 + 33)}^4 - {(40 \cdot 384140 + 39)}^4 \equiv 0 \pmod {40^4}.$$

Let's expand those numbers for the time being, and also move the $\equiv$ sign to the middle:

$$20615673^4 \equiv 15365639^4 \pmod {40^4}.$$

Now, since we're working modulo $40^4$, we can disregard any multiple of $40^4$ in the above expression. So this identity reduces down to:

$$135673^4 \equiv 5639^4 \pmod {40^4}.$$

The number $40$ factors into prime powers as $8 \cdot 5$, so by the Chinese remainder theorem, the above identity is equivalent to these two identities:

$$135673^4 \equiv 5639^4 \pmod {8^4};\\ 135673^4 \equiv 5639^4 \pmod {5^4}.$$

Again, we can disregard multiples of $8^4$ and $5^4$, so these identities reduce further down to:

$$505^4 \equiv 1543^4 \pmod {8^4};\\ 48^4 \equiv 14^4 \pmod {5^4}.$$

I don't know how to explain these last two identities, but hopefully they're easier to explain than your original identity.

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  • $\begingroup$ You are right ! A very good analysis. Let us see if these last two identities tell us anything about the original question !! $\endgroup$ – Derek Jun 26 '17 at 23:50

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