If positive integers $a$, $b$, $c$ satisfy $\frac1{a^2}+\frac1{b^2}=\frac1{c^2}$, then the sum of all values of $a\leq 100$ is ...

Hint 1) $a$ must be a multiple of $5$.

Hint 2) You need only consider multiples of the $(3,4,5)$ and $(5,12,13)$ triangles.


Let $(a,b)=d$

WLOG $\dfrac aA=\dfrac bB=d\implies(A,B)=1$

$$c^2(A^2+B^2)=A^2B^2$$

$$\implies\left(\dfrac{AB}c\right)^2=A^2+B^2$$ which is an integer

So, $c|AB, c=CAB$(say)

$$\implies A^2+B^2=C^2$$

As $(A,B)=1$

WLOG $A,B\in[m^2-n^2,2mn],C=m^2+n^2$

So, we need $2mn$ to divide $100\iff mn|50$

or $m^2-n^2$ to divide $100$

and of course $(2mn,m^2-n^2)=1$

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Elementary Number Theory

Algebra Precalculus

Contest Math

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