# find the smallest positive integer that can be written as a sum of two squares in three different ways

(HMMT 2000 Guts Round #27). Find the smallest positive integer that can be written as a sum of two squares in exactly three different ways.

By using a computer program, I found the number $$325.$$ It can only be written as $$15^2 + 10^2, 1^2+18^2,6^2+17^2.$$ One can verify this by observing that if $$a^2 + b^2 = 325,$$ then we may assume WLOG that $$a\leq b$$ so that $$2a^2 \leq 325\Rightarrow a \leq 12.$$ Thus it suffices to find the values of a that are at most 12 and such that $$325-a^2$$ is a perfect square. It could be useful to list the first few perfect squares: $$0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324.$$ Now suppose for a contradiction that some number that's at most $$324$$ can be expressed as a sum of two squares in three different ways. Note that all the perfect squares listed can be expressed as a sum of two squares in at most two different ways. I know that if $$x^2 + y^2=z^2$$ is a primitive Pythagorean triple of positive integers with x odd, then there exists integers $$u>v>0$$ of different parity so that $$x=u^2-v^2, y = 2uv, z=u^2+v^2$$. So one could use this theorem to find some primitive Pythagorean triples with $$z\leq 18$$. For instance, we can choose $$u=3,v=2$$ to get $$x=5, y=12,z=13$$. By checking values of $$v$$ that are at most 2, we see that the only other primitive triples with $$z\leq 18$$ are $$(x,y,z) = (15,8,17)$$ and $$(3,4,5).$$ Thus since any Pythagorean triple is a "multiple" of a primitive triple, the only Pythagorean triples with $$z\leq 18$$ are $$(3,4,5),(9,12,15), (5,12,13),(15,8,17).$$ Each value of $$z$$ leads to a unique pair $$(x,y)$$ and so the claim that all perfect squares that are at most 324 can only be expressed as the sum of two squares in at most 2 ways holds. So we just need to consider non-perfect squares, though I'm not sure of an approach that isn't computationally intensive.

• – lulu
Dec 16, 2022 at 0:41

Let $$n$$ be the smallest integer that can be written as a sum of two squares in exactly three different ways. Then $$n=x_1^2+x_2^2=x_3^2+x_4^2=x_5^2+x_6^2,$$ for some nonnegative integers $$x_i$$. Consider the prime factorization of $$n$$:

If $$n$$ is divisible by a prime $$p\equiv3\pmod{4}$$, then also the $$x_i$$ are all divisible by $$p$$. But then dividing the $$x_i$$ by $$p$$ shows that $$\frac{n}{p^2}$$ is a sum of two squares in exactly three different ways, contradicting the minimality of $$n$$.

Similarly, if $$n$$ is divisible by $$4$$ then reducing mod $$4$$ shows that all the $$x_i$$ are even. But then dividing the $$x_i$$ by $$2$$ shows that $$\frac{n}{4}$$ is a sum of two squares in exactly three different ways, contradicting the minimality of $$n$$.

If $$n\equiv2\pmod{4}$$ then reducing mod $$4$$ shows that all the $$x_i$$ are odd. But then the identity $$\left(\frac{x_1+x_2}{2}\right)^2+\left(\frac{x_1-x_2}{2}\right)^2=\frac{x_1^2+x_2^2}{2}=\frac{n}{2},$$ shows that $$\frac{n}{2}$$ is a sum of two squares in exactly three different ways, contradicting the minimality of $$n$$.

This shows that all prime factors of $$n$$ are congruent to $$1$$ modulo $$4$$. Of course if $$n$$ is prime then it can be written as a sum of squares in only one way. If $$n$$ is a product of two primes then it can be written as a sum of squares in at most two ways. So $$n$$ must be a product of at least three primes congruent to $$1$$ mod $$4$$.

The smallest such number is $$5^3=125$$. If $$125=x_i^2+x_j^2$$ with $$x_i>x_j$$ then $$x_i\geq8$$ and of course $$x_i\leq11$$. It is then easy to check that $$125$$ can not be written as a sum of two squares in exactly three ways.

The next such number is $$5^2\times13=325$$. If $$325=x_i^2+x_j^2$$ with $$x_i>x_j$$ then $$x_i\geq13$$, and of course $$x_i\leq18$$. It is then easy to check that $$325$$ can be written as a sum of two squares in exactly three ways.