$ Lemma: x^{2} + y^{2} = z^{2}$ ,$ (p,q,r)$ is permutation of $ (x,y,z)$ $ \implies$ $ 3.4.5|pqr$ .$ \implies$ $ 3|n,4|n,5|n$ $ 1 = d_{1} < 2 < 3 < 4 < 5 < 6 < d_{7}$ $ \implies$ $ 7\leq d_{7}\leq 10$ $ \implies$ $ i)d_{7} = 7$ $ \implies$, $ d_{15} = a,d_{16} = b$ $ \implies$ $ (b - a)(b + a) = 49$ $ \implies$ only solution is $ (25,24)$ $ 1 = d_{1} < 2 < 3 < 4 < 5 < 6 < 7 < 8 < d_{9} < d_{10} < d_{11} < ... < 24 < 25$ we need $ 6$ number. $ 10,12,14,15,20,21$ are divisors of $ n$ $ \implies$ $ 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 10 < 12 < 14 < 15 < 20 < 21 < 24 < 25 < d_{17}$ $ d_{17} = 28$ because $ 13\nmid n,9\nmid n$ $ \implies$ $ 26\nmid n,27\nmid n$ and $ 7|n,4|n$ $ \implies$ $ 28|n$ similarly $ d_{7} = 8,9,10$ $ \Box$

cnyd wrote: $ Lemma: x^{2} + y^{2} = z^{2}$ ,$ (p,q,r)$ is permutation of $ (x,y,z)$ $ \implies$ $ 3.4.5|pqr$ .$ \implies$ $ 3|n,4|n,5|n$ Would anyone prove this lemma'?

Vulch wrote: cnyd wrote: $ Lemma: x^{2} + y^{2} = z^{2}$ ,$ (p,q,r)$ is permutation of $ (x,y,z)$ $ \implies$ $ 3.4.5|pqr$ .$ \implies$ $ 3|n,4|n,5|n$ Would anyone prove this lemma'? $t^2 \equiv 0,1(mod \ 3)$ If $3\nmid pqr$, then $1+1\equiv 1(mod \ 3)$, contradiction the case of $mod\ 4, mod\ 5$ is the same way

Why only $3,4,5$ has been chosen as Pythagorean triples for the equation $x^2 +y^2=z^2?$ Can't be other Pythagorean triples satisfying the Pythagoras equation $x^2 +y^2=z^2?$ Why $d_7=7?$ Since,it's given that $ d_1,d_2 \ldots, d_r$ be the positive divisors of $ n.$ So,does $7$ is divisor of $60?$

Vulch wrote: Why only $3,4,5$ has been chosen as Pythagorean triples for the equation $x^2 +y^2=z^2?$ Can't be other Pythagorean triples satisfying the Pythagoras equation $x^2 +y^2=z^2?$ Why $d_7=7?$ Since,it's given that $ d_1,d_2 \ldots, d_r$ be the positive divisors of $ n.$ So,does $7$ is divisor of $60?$ No,the lemma means $60|xyz$ and from $7\leq d_7\leq 10$,we need to discuss the value if $d_7$.

cnyd wrote: we need $6$ number. $ 10,12,14,15,20,21$ are divisors of $n.$ How did 'cnyd' get it?

Vulch wrote: cnyd wrote: we need $6$ number. $ 10,12,14,15,20,21$ are divisors of $n.$ How did 'cnyd' get it? The six divisors we're talking about are $d_9,\ldots,d_{14}$. Divisors $10,12,15,20$ come from $60|n$ (deduced from $d_7^2+d_{15}^2=d_{16}^2$). Divisors $14,21$ are from the assumption in case (i) that $d_7=7$, along with factors of $2,3$ from $60|n$.