Question Number 149567 by mathdanisur last updated on 06/Aug/21

if q is prime number fixed, then solve for natural numbers the equation: (1/q) = (1/x) + (1/y) - (1/z)

Commented byRasheed.Sindhi last updated on 07/Aug/21

My answer is too lengthy. The question is waiting for a better solution.

Commented bymathdanisur last updated on 07/Aug/21

Dear Ser We hawe: (1/n) = (1/(2(n-1))) + (1/(2(n+1))) - (1/((n-1)(n+1))) ⇒ n=q .... Ser, this true or wrong.?

Commented byRasheed.Sindhi last updated on 07/Aug/21

(1/(2(q−1)))+(1/(2(q+1)))−(1/((q−1)(q+1))) =((q+1+q−1−2)/(2(q−1)(q+1)))=((2q−2)/(2(q−1)(q+1))) =((2(q−1))/(2(q−1)(q+1)))=(1/(q+1))≠(1/q) Not correct ser.

Commented bymathdanisur last updated on 07/Aug/21

Sorry Ser, ... (1/(n(n-1)(n+1)))

Commented byRasheed.Sindhi last updated on 07/Aug/21

(1/(2(n-1))) + (1/(2(n+1))) - (1/(n(n-1)(n+1))) ((n(n+1)+n(n−1)−2)/(2n(n-1)(n+1))) ((n^2 +n+n^2 −n−2)/(2n(n-1)(n+1)))=((2n^2 −2)/(2n(n-1)(n+1))) ((2(n−1)(n+1))/(2n(n-1)(n+1)))=(1/n) ✓ It means that this is the solution.

Answered by Rasheed.Sindhi last updated on 07/Aug/21

(1/q) = (1/x) + (1/y) - (1/z) ;q∈P∧ x,y,z∈N ((xyz)/q)=−xy+yz+zx................(A) q∣x ∨ q∣y ∨ q∣z 3 cases: { ((C(1):x=mq)),((C(2):y=mq)),((C(3):z=mq)) :} C(1): x=mq ; m∈{1,2,3,...} (A)⇒myz=−mqy+yz+mqz yz(m−1)=mq(z−y) yz=((mq(z−y))/(m−1))................(B) (m−1)∣m ∨ (m−1)∣q ∨ (m−1)∣(z−y) : { ((C(1a):(m−1)∣m⇒m−1=1⇒m=2)),((C(1b):(m−1)∣q⇒m=2 or m=q+1)),((C(1c):(m−1)∣(z−y)⇒y≡z(mod m−1))) :} C(1a):m=2 (B)⇒yz=2q(z−y)⇒y∈E ∨ z∈E : { (((∗)y=2u,z=2v⇒4uv=4q(v−u))),(((∗∗)y=2u⇒2uz=2q(z−2u))),(((∗∗∗)z=2u⇒2uy=2q(2u−y))) :} (∗)uv=q(v−u): { ((qk.ql=q(ql−qk)⇒kl=l−k^• )),((qk.v=q(v−qk)⇒kv=v−qk^(••) )),((u.qk=q(qk−u)⇒uk=(qk−u))) :} ^• lk−l+k=0 l(k−1)=−k l=(k/(1−k))⇒(1−k)∣k ⇒1−k=1⇒k=2⇒u=qk=2q ⇒y=2u=4q l=(2/(1−2))=−2⇒v=ql=−2q ⇒z=2v=2(−2q)=−4q∉N x=mq=2q (x,y,z)=(2q,4q,−4q) if x,y,z∈Z ^(••) kv=v−qk qk=v−kv=v(1−k) qk=v(1−k) q=−((v(k−1))/k) k∤(k−1)⇒k∣v v=kt⇒q=−t(k−1) ∵q∈P⇒k−1=−1⇒k=0⇒q=t ⇒v=qk⇒z=2v=2qk u=qk⇒y=2u=2qk Continue...

Commented byRasheed.Sindhi last updated on 07/Aug/21

If we allow x,y,z∈Z (x,y,z)=(2q,4q,−4q) is a general solution. Verification: (1/q) = (1/x) + (1/y) - (1/z) (1/q) = (1/(2q)) + (1/(4q)) - (1/(−4q))=((2+1+1)/(4q))=(1/q) Too lengthy to continue

Commented bymathdanisur last updated on 07/Aug/21

Ser, thank you cool

Answered by Rasheed.Sindhi last updated on 06/Aug/21

(1/q)=(1/x)+(1/y)−(1/z) ;q∈P ∧ x,y,z∈N (It′s obvious that x & y are exchangeable.) ((xyz)/q)=−xy+yz+zx q∣x ∨ q∣y ∨ q∣z case1:x=mq myz=−mqy+yz+mqz yz(m−1)=−mq(y−z) yz=((mq(z−y))/(m−1)) { (((m−1) ∣ m⇒m=2)),(((m−1) ∣ q⇒m−1=1 or m−1=q⇒ { ((m=2)),((m=q+1)) :})),(((m−1)∣ (z−y))) :} subcase1:m=2 yz=2q(z−y) At least one of y , z is even. ((yz)/(z−y))=2q (a): y,z∈E:y=2u,z=2v ((2u.2v)/(2v−2u))=2q ((uv)/(v−u))=q.......(∗) { ((v−u=1)),(((v−u)∣uv)) :} v−u=1⇒v=u+1 u(u+1)=q⇒u=1⇒q=2 (∗)⇒(v/(v−1))=2⇒v=2v−2⇒v=2 y=2u=2(1)=2,z=2v=2(2)=4 x=mq=2×2=4 (x,y,z)=(4,2,4) for q=2 Continue

Commented bymathdanisur last updated on 06/Aug/21

Ser, thank you

Commented byRasheed.Sindhi last updated on 07/Aug/21

Two (obvious) general solutions: x=z ∧ y=q y=z ∧ x=q