The positive integers $a, b$ and $c$ satisfy $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}$.

The sum of all possible values of $a \leq 100$ is
(A) 315
(B) 615
(C) 680
(D) 555
(E) 620
The first challenge in this problem is to find one set of numbers $a, b, c$ that actually works. Since this looks a bit similar to the Pythagorean Theorem, we can start with $3^{2}+4^{2}=5^{2}$ and try to manipulate this.

If we divide both sides by the least common multiple of $3^{2}, 4^{2}$ and $5^{2}$, which is $(3 \times 4 \times 5)^{2}=60^{2}$, we then obtain $\frac{3^{2}}{60^{2}}+\frac{4^{2}}{60^{2}}=\frac{5^{2}}{60^{2}}$ or $\frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$.

This gives us two possible triples: $(a, b, c)=(20,15,12)$ and $(a, b, c)=(15,20,12)$ (so two possible values for $a$ so far). How can we get more? We can multiply the equation $\frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$ by reciprocals of perfect
squares.

Multiplying by $\frac{1}{2^{2}}$, we get $\frac{1}{40^{2}}+\frac{1}{30^{2}}=\frac{1}{24^{2}}$.

Multiplying by $\frac{1}{3^{2}}$, we get $\frac{1}{60^{2}}+\frac{1}{45^{2}}=\frac{1}{36^{2}}$.

Multiplying by $\frac{1}{4^{2}}$, we get $\frac{1}{80^{2}}+\frac{1}{60^{2}}=\frac{1}{48^{2}}$.

Multiplying by $\frac{1}{5^{2}}$, we get $\frac{1}{100^{2}}+\frac{1}{75^{2}}=\frac{1}{60^{2}}$.

Multiplying by $\frac{1}{6^{2}}$, we get $\frac{1}{120^{2}}+\frac{1}{90^{2}}=\frac{1}{72^{2}}$.

Multiplying by $\frac{1}{7^{2}}$, we get $\frac{1}{140^{2}}+\frac{1}{105^{2}}=\frac{1}{84^{2}}$.

At this point, the strategy will no longer work, since we are only looking for values of $a \leq 100$. So far, the possible values of $a$ are (from looking at each denominator of the left side of each the equations here): $20,15,40,30,60,45,80,100,75,90$. (Notice that 60 doesn't appear twice in the list!) The sum of these numbers is 555 .

Can we find more starting equations by beginning with a different Pythagorean triple? If we start with $5^{2}+12^{2}=13^{2}$ and divide both sides by the least common multiple of $5^{2}, 12^{2}$ and $13^{2}$ (ie. $(5 \times 12 \times 13)^{2}=780^{2}$ ), we get $\frac{1}{156^{2}}+\frac{1}{65^{2}}=\frac{1}{60^{2}}$ which gives us 65 as another possible value of $a$.

Therefore, our running total for values of $a$ is $555+65=620$.

We can't generate more possible values for $a$ using $\frac{1}{156^{2}}+\frac{1}{65^{2}}=\frac{1}{60^{2}}$ since multiplying both sides
by the reciprocal of any perfect square will make both of $a$ and $b$ at least 130, so bigger than 100 .

Can we use $6^{2}+8^{2}=10^{2}$ ? Here, the least common multiple of $6^{2}, 8^{2}$ and $10^{2}$ is $120^{2}$, and dividing by $120^{2}$ gives us $\frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$, which we have already used.

Can we use any other Pythagorean triple? No, since any other Pythagorean triple is at least as big as 7-24-25, and so the smallest possible denominator that we will get on the left side by using this technique is $(7 \times 25)^{2}=175^{2}$, which would give an $a$ larger than 100 .

Also, any triple $(a, b, c)$ that actually works does come from a Pythagorean triple, since we can multiply both sides of $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}$ by $(a b c)^{2}$ to get $(b c)^{2}+(a c)^{2}=(a b)^{2}$.

So every possible triple $(a, b, c)$ comes from a Pythagorean triple, and no Pythagorean triples give any more allowable values of $a$, so we have found them all. Therefore, the sum of all possible values of $a \leq 100$ is 620 .

ANswer: (E)