# Is-there-a-solution-to-i-a-n-i-i-a-n-i-n-a-Z-

Question Number 3193 by Filup last updated on 07/Dec/15
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}: \\$$$$\underset{{i}={a}} {\overset{{n}} {\sum}}{i}=\underset{{i}={a}} {\overset{{n}} {\prod}}{i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({n},\:{a}\right)\in\mathbb{Z} \\$$
Commented by Filup last updated on 07/Dec/15
$$\underset{{i}={a}} {\overset{{n}} {\sum}}{i}=\frac{\mathrm{1}}{\mathrm{2}}{n}\left({a}+{n}\right) \\$$$$\underset{{i}={a}} {\overset{{n}} {\prod}}{i}=\frac{{n}!}{\left({a}−\mathrm{1}\right)!} \\$$$$\\$$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}{n}\left({a}+{n}\right)=\frac{{n}!}{\left({a}−\mathrm{1}\right)!} \\$$
Commented by prakash jain last updated on 07/Dec/15
$${n}={a}\: \\$$$${n}={a}\:\underset{{i}={a}} {\overset{{n}} {\prod}}{i}={a}=\underset{{i}={a}} {\overset{{n}} {\sum}}{i} \\$$$${your}\:{formula}\:{for}\:\underset{{i}={a}} {\overset{{n}} {\sum}}{i}\:\mathrm{is}\:\mathrm{incorrect}.\:\mathrm{it}\:\mathrm{should} \\$$$$\mathrm{be}=\frac{{n}−{a}+\mathrm{1}}{\mathrm{2}}\left({a}+{n}\right) \\$$$$\mathrm{If}\:{n}>{a}\:{and}\:{a}>\mathrm{1} \\$$$${a}\left({a}+\mathrm{1}\right)\neq{a}+{a}+\mathrm{1} \\$$$${so}\:{no}\:{solutions}\:{for}\:{n}>{a} \\$$$${n}<{a}\:\mathrm{LHS}=\mathrm{0}=\mathrm{RHS} \\$$
Commented by Filup last updated on 07/Dec/15
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{correction}! \\$$$$\mathrm{W}{hat}\:\mathrm{if}\:\:\:\:\:{a},{n}\in\mathbb{R}? \\$$$${Same}\:{result}? \\$$
Commented by Filup last updated on 07/Dec/15
$$\mathrm{What}\:\mathrm{if} \\$$$${a}+\left({a}+\mathrm{1}\right)+…+\left({n}\right)={a}\left({a}+\mathrm{1}\right)…{n} \\$$$${for}\:\:\left({a},\:{n}\right)\in\mathbb{R} \\$$
Commented by prakash jain last updated on 07/Dec/15
$$\mathrm{If}\:{a}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{real}.\:\mathrm{Then}\:\mathrm{you}\:\mathrm{would}\:\mathrm{get} \\$$$$\mathrm{a}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{degree}\:{n}−{a}+\mathrm{1}\:\mathrm{with}\:{n}−{a}+\mathrm{1} \\$$$$\mathrm{solutions}.\:\mathrm{Some}\:\mathrm{of}\:\mathrm{them}\:\mathrm{will}\:\mathrm{be}\:\mathrm{complex}. \\$$