MHB Counting Integers of a Specific Form

[lfdahl]

Let $a, b \le 2015$ be positive integers. What is the number of integers of the form:

\[ \frac{a^4+b^4}{625}? \]

 

[Greg]

Let $a, b \le 2015$ be positive integers. What is the number of integers of the form:

\[ \frac{a^4+b^4}{625}? \]

Spoiler

It suffices that a and b are multiples of 5 between 1 and 2015. 2015/5 = 403, hence the desired number is 403$^2$ = 162409.

 

[HallsofIvy]

Spoiler

Greg1313 is using the fact that 625= 5^4.

 

[lfdahl]

Spoiler

It suffices that a and b are multiples of 5 between 1 and 2015. 2015/5 = 403, hence the desired number is 403$^2$ = 162409.

Thankyou, greg1313, for your answer.

Spoiler

Your statement, that $a$ and $b$ are multiples of 5 is true, - why?

 

[Greg]

Thankyou, greg1313, for your answer.

Spoiler

Your statement, that $a$ and $b$ are multiples of 5 is true, - why?

[sp]5$^4$ = 625[/sp]

 

[lfdahl]

[sp]5$^4$ = 625[/sp]

Hi, again, greg1313:

Spoiler

Your argument is insufficient. Maybe there exist $a$ and $b$ such, that:

\[a \not\equiv 0\: \: \wedge b\not\equiv 0 \: \: (mod \: \: 5) \\\\ and \\\\\frac{a^4+b^4}{5^4}\in \mathbb{N}\]

You need to prove, why we cannot choose $a$ and $b$ in this way ...

 

[Greg]

Hi, again, greg1313:

Spoiler

Your argument is insufficient. Maybe there exist $a$ and $b$ such, that:

\[a \not\equiv 0\: \: \wedge b\not\equiv 0 \: \: (mod \: \: 5) \\\\ and \\\\\frac{a^4+b^4}{5^4}\in \mathbb{N}\]

You need to prove, why we cannot choose $a$ and $b$ in this way ...

Spoiler

Yes, I'm afraid I must agree. However, by Fermat's little theorem,

$$n^4\equiv1\pmod5$$

if $n$ is not divisible by $5$, and $0$ otherwise. Hence,

$$a^4+b^4\equiv0,1,2\pmod5$$

with $0$ occurring iff $a$ and $b$ are both divisible by $5$.
It follows that $a^4$ and $b^4$ must both be divisible by $5$
and since $5^4=625,\,a^4+b^4$ is divisible by $625$ iff $a$
and $b$ are divisible by $5$.

Thank you for your assistance. :)

 

[lfdahl]

Well done! Thankyou, greg1313!