Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that there is some natural number $a$ and $b$. Now we perform $c = a^2 + b^2$. This time, c is even.

Will this $c$ only have one possible pair of $a$ and $b$?

edit: what happens if c is odd number?

share|improve this question
    
Have you seen this answer? It gives you results related to the number of ways of writing a given $n$ as a sum of two squares, subject to sundry conditions. –  Arturo Magidin Jul 11 '12 at 15:11
add comment

2 Answers

up vote 3 down vote accepted

Not necessarily. For example, note that $50=1^2+7^2=5^2+5^2$, and $130=3^2+11^2=7^2+9^2$. For an even number with more than two representations, try $650$.

We can produce odd numbers with several representations as a sum of two squares by taking a product of several primes of the form $4k+1$. To get even numbers with multiple representations, take an odd number that has multiple representations, and multiply by a power of $2$.

To help you produce your own examples, the following identity, often called the Brahmagupta Identity, is quite useful: $$(a^2+b^2)(x^2+y^2)=(ax\pm by)^2 +(ay\mp bx)^2.$$

share|improve this answer
add comment

If a,b are both even or both odd, c is even.

c will be odd iff a and b are opposite parity.

Let a=2A and b=2B+1, then c≡1(mod 4).

So if c≡3≡-1(mod 4), there will be no solution.


In other way, we can take (a,b)=d, then $d^2|c$ , let $C^2=d^2.c$

Let $\frac{a}{A}=\frac{b}{B}=d$, clearly (A,B)=1

Then, A & B are both not even.

The case of A,B being opposite parity has been dealt above.

If A & B are both odd, let A=2m+1, B=2n+1.

$A^2+B^2=(2m+1)^2+(2n+1)^2=2(p^2+q^2)$ (say)=$(p+q)^2+(p-q)^2$

Then p=m+n+1, q=m-n and p+q=2m+1 is odd=> p & q are of opposite parity.

So, the problem boils down to finding A,B such that (A,B)=1 and A,B are of opposite parity.

Clearly, C≡1(mod 4) is a necessary condition for solubility.

Proof of sufficiency

Brahmagupta Identity can used to prove that if C is a product of n primes of the form 4r+1, it can be represented as the sum of two squares in $2^{n-1}$ ways.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.