Exercise 3 on page 5 and exercise 7 on page 6 in Koblitz's Introduction to modular forms.

Question

I want to prove $1$ cannot be a congruent number, by using the fact that if it were congruent then the equation $x^4-y^4=u^2$ would have a solution in integers with $u$ being odd.

I proved this last claim, but I don't see how to assert from this claim a contradiction.

Can anyone provide me with hints? I assume that I am missing some result in number theory.

As for exercise 7a, it asks me to show that condition B of Tunnel's theorem always holds for if $n$ is congruent to $5$ or $7$ modulo $8$.

Tunnell's theorem states the following:

Let $n$ be an odd squarefree natural number. Consider the two conditions: $(A)$ $n$ is a congruent number;

$(B)$ the number of triples of integers $(x,y,z)$ satisfying $2x^2+y^2+8z^2 = n$ is equal to twice the number of triples satisfying $2x^2+y^2+32z^2 = n $, then $(A)$ implies $(B)$ and if a weak form of Birch-Swinnerton-Dyer conjecture is true, then $(B)$ also implies $(A)$.

Now I am stuck on showing that condition $(B)$ is satisfied by $n = 8m+5$. How to show that the number of triples $(x,y,z)$ satisfying $2x^2+y^2+8z^2 = 5+8m$ is twice the number of triples that satisfy: $2x^2+y^2+32z^2 = 8m+5$; I thought of moving the $8m$ to the LHS. I thought of the fact that $z \mapsto \pm 2z$ changes one equation to another, but not sure how to continue from there.

Any tips?

Thanks in advance.

P.S A congruent number is a rational number which is the area of some right triangle with rational sides, i.e $r$ is called a congruent number if there exist rational numbers $X,Y,Z$ s.t $X^2+Y^2=Z^2 , XY/2=r$.

Answer 1

Just reading a text and not trying to actually compute anything might not be a good advice for learning math, especially number theory. Prof. Koblitz have written the question in a rather obscure manner, but trying to actually put some numbers in the equation reveals the trick.

I claim that when $m=5,7 \bmod(8)$, there are NO solutions for the congruence. By looking $\bmod (2)$ one figures out that $y$ is odd, hence $y^2\equiv 1 \bmod(8)$ therefore one left with $2x^2\equiv 4,6 \bmod(8)$, which have no solutions.