# Divisors of $25^2+98^2$

How many divisors does $25^2+98^2$ have?

My Attempt:

Calculator is not allowed but using calculator I found $193\times53$ that means $8$ divisors and that $4$ of them are positive.

• One way that may allow to arrive at the result without finding the divisors:Verify that the number is $<23^3$ and not divisible by primes $<23$. Then it can only be prime (4) or the square of a prime (6) or the product of two distinct primes (8). Square of prime can be excluded because the number is $\eqiuv 5\pmod 8$. May 29, 2016 at 9:31
• Just for fun: $25+98i=(7+2i)(7+12i)$ gives $25^2+98^2=|25+98i|^2=|7+2i|^2 |7+12i|^2=53 \cdot 193$.
– lhf
Jun 3, 2016 at 13:14

The trick here is the useful identity $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$. In this case we have $a=5,b=7$, so we get the factorisation $(25+70+98)(25-70+98)=193\cdot53$.

It is now easy to check that 53,193 are prime, so there are as you say 4 divisors: 1, 53, 193 and the number itself.

It is usual in this kind of question only to count the positive divisors. Obviously there are another 4 negative divisors.

HINT:

$a^2+b^2=(a+b)^2-2ab$

Hence:

$25^2+98^2=$

$(25+98)^2-2\cdot25\cdot98=$

$123^2-4900=$

$123^2-70^2=$

$(123-70)\cdot(127+70)=$

$53\cdot193$

$25^2+98^2=123^2-2 \times 98 \times 25=123^2-70^2=193 \times 53$

I found this answer using almagest's answer thanks a lot for help.