By Sophie-Germain, $(n^4+4)=(n^2+2n+2)(n^2-2n+2)$ Note also that $(n+2)^2-2(n+2)+2=n^2+4n+4-2n-4+2=n^2+2n+2,$ so we have a telescoping product. Then, our expression is equal to $$\frac{1^2-2+2}{3^2+2*3+2}\cdot\frac{5^2-2*5+2}{7^2+2*7+2}\cdots\frac{73^2-2*73+2}{75^2+2*75+2},$$which gives $\frac{1^2-2+2}{75^2+2*75+2}=\frac{1}{5777}.$

awesomeclaw wrote: By Sophie-Germain, $(n^4+4)=(n^2+2n+2)(n^2-2n+2)$ Note also that $(n+2)^2-2(n+2)+2=n^2+4n+4-2n-4+2=n^2+2n+2,$ so we have a telescoping product. Then, our expression is equal to $$\frac{1^2-2+2}{3^2+2*3+2}\cdot\frac{5^2-2*5+2}{7^2+2*7+2}\cdots\frac{73^2-2*73+2}{75^2+2*75+2},$$which gives $\frac{1^2-2+2}{75^2+2*75+2}=\frac{1}{5777}.$ Sadly, the answer is wrong. You made a mistake in considering some of the terms. Also, technically () Sophie Germain actually states:

$$a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$$Proof

We want: $$\prod_{n=1}^{18} \frac {(1+4n)^4+4}{(3+4n)^4+4}$$Applying that with $a=1, 2, 3 \cdots$ and $b=1$ gives: $$\frac{(1^4+4)\cdot (5^4+4)\cdot (9^4+4)\cdot ... (69^4+4)\cdot(73^4+4)}{(3^4+4)\cdot (7^4+4)\cdot (11^4+4)\cdot ... (71^4+4)\cdot(75^4+4)} $$$$\implies \frac {5 \cdot 37 \cdot 17 \cdot 101 \cdot 65 \cdot 145 \cdot 197 }{17 \cdot 101 \cdot 5 \cdot 65 \cdot 145 \cdot 101 \cdots }$$$$\implies \frac {\cancel 5 \cdot 37 \cdot \cancel {17} \cdot \cancel {101} \cdot \cancel {65} \cdot \cancel {145} \cdot \cancel {197} }{\cancel {17} \cdot \cancel {101} \cdot \cancel {5} \cdot \cancel {65} \cdot \cancel {145} \cdot \cancel {101} \cdots }.$$ This means everything in the numerator and denominator will cancel out except for the $37$. We look at the last few terms to see what they give: $$\frac {(69^4+4)(73^4+4)}{(71^4+4)(75^4+4)}=\frac {4625*4901*5477*5185}{4901 \cdot 5185 \cdot 5777\cdot 5477}$$$$\implies \frac {\cancel {4625} \cdot \cancel {4901} \cdot \cancel {5477} \cdot \cancel {5185}}{\cancel {4901} \cdot \cancel {5185} \cdot 5777 \cdot \cancel {5477}}$$ And the $5777$ does not cancel. We are left with: $$\boxed {\frac {37}{5777}}$$ but wolfram says $\frac {17}{5777}$. Where did I go wrong? http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=393950e05a927d5e57b7491d2e9ceeca&title=Math%20Help%20Boards%3A%20Product%20Calculator&theme=blue

$$\prod_{n=1}^{18} \frac {(1+4n)^4+4}{(3+4n)^4+4} = \frac{\prod_{n=1}^{18} ((1+4n)^2+4)}{\prod_{n=1}^{18} ((3+4n)^2+4)}=\frac{\prod_{n=1}^{18} (16n^2+1)(16n^2+16n+5)}{\prod_{n=1}^{18} (16n^2+16n+5)(16n^2+32n+17)}=\frac{\prod_{n=1}^{18} (16n^2+1)}{\prod_{n=1}^{18} (16(n+1)^2+1)}=\frac{16+1}{16\cdot 19^2+1}=\frac{17}{5777}$$

hinna wrote: awesomeclaw wrote: By Sophie-Germain, $(n^4+4)=(n^2+2n+2)(n^2-2n+2)$ Note also that $(n+2)^2-2(n+2)+2=n^2+4n+4-2n-4+2=n^2+2n+2,$ so we have a telescoping product. Then, our expression is equal to $$\frac{1^2-2+2}{3^2+2*3+2}\cdot\frac{5^2-2*5+2}{7^2+2*7+2}\cdots\frac{73^2-2*73+2}{75^2+2*75+2},$$which gives $\frac{1^2-2+2}{75^2+2*75+2}=\frac{1}{5777}.$ Sadly, the answer is wrong. You made a mistake in considering some of the terms. Also, technically () Sophie Germain actually states:

$$a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$$Proof

We want: $$\prod_{n=1}^{18} \frac {(1+4n)^4+4}{(3+4n)^4+4}$$Applying that with $a=1, 2, 3 \cdots$ and $b=1$ gives: $$\frac{(1^4+4)\cdot (5^4+4)\cdot (9^4+4)\cdot ... (69^4+4)\cdot(73^4+4)}{(3^4+4)\cdot (7^4+4)\cdot (11^4+4)\cdot ... (71^4+4)\cdot(75^4+4)} $$$$\implies \frac {5 \cdot 37 \cdot 17 \cdot 101 \cdot 65 \cdot 145 \cdot 197 }{17 \cdot 101 \cdot 5 \cdot 65 \cdot 145 \cdot 101 \cdots }$$$$\implies \frac {\cancel 5 \cdot 37 \cdot \cancel {17} \cdot \cancel {101} \cdot \cancel {65} \cdot \cancel {145} \cdot \cancel {197} }{\cancel {17} \cdot \cancel {101} \cdot \cancel {5} \cdot \cancel {65} \cdot \cancel {145} \cdot \cancel {101} \cdots }.$$ This means everything in the numerator and denominator will cancel out except for the $37$. We look at the last few terms to see what they give: $$\frac {(69^4+4)(73^4+4)}{(71^4+4)(75^4+4)}=\frac {4625*4901*5477*5185}{4901 \cdot 5185 \cdot 5777\cdot 5477}$$$$\implies \frac {\cancel {4625} \cdot \cancel {4901} \cdot \cancel {5477} \cdot \cancel {5185}}{\cancel {4901} \cdot \cancel {5185} \cdot 5777 \cdot \cancel {5477}}$$ And the $5777$ does not cancel. We are left with: $$\boxed {\frac {37}{5777}}$$ but wolfram says $\frac {17}{5777}$. Where did I go wrong? http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=393950e05a927d5e57b7491d2e9ceeca&title=Math%20Help%20Boards%3A%20Product%20Calculator&theme=blue Actually, he's not wrong - note that for $n=1$, $4n+1=5$, so your product should start from 0 and not from 1. Also, I think you got the denominator wrong, since $7^4+4 = 37*65$.