Question Number 100178 by bobhans last updated on 25/Jun/20

what is the number of ordered pairs of positif integers (x,y) that satisfy x^2 +y^2 −xy=37

Answered by 1549442205 last updated on 25/Jun/20

x^2 −xy+y^2 −37=0.We look at it like a quadratic equation respect with x.Then Δ=y^2 −4y^2 +148=−3y^2 +148.The above eqs.always has roots ,so Δ=−3y^2 +148≥0 ⇒y^2 ≤((148)/3).Since y∈N^(∗ ) ,y∈{1,2,...,7}. i)If y∈{1,2,5,6}then Δ∈{145,136,73,40} aren′t perfect numbers,so are rejected ii)for y=3⇒Δ=121=11^2 ⇒x=((3+11)/2)=7(we reject negative root) we get the root (x,y)=(7;3) iii)for y=4 ⇒Δ=100=10^2 ⇒x=((4+10)/2)=7 we get the root (x;y)=(7;4) iv)for y=7 ⇒Δ=1⇒x=((7±1)/2)⇒x∈{4;3} we get two roots (x;y)∈{(4;7);(3;7)} Thus ,the given equation has four roots (x;y)∈{(7;3);(7;4);(3;7);(4;7)}

Commented bybemath last updated on 25/Jun/20

how with (4,−3),(−4,3),(3,−4),(−3,4)?

Commented by1549442205 last updated on 25/Jun/20

your problem require to find the pairs of positive integers

Answered by Rasheed.Sindhi last updated on 25/Jun/20

A Novel Method_(−) x^2 +y^2 −xy=37 (x−y)^2 =37−xy............A (x+y)^2 =37+3xy............B x,y∈Z^+ ∧ 37−xy is perfect square 37−xy=36,25,16,9,4,1(possible values) xy=1,12,21,28,33,36 From above values of xy acceptable values are those for which 37+3xy is also perfect square are 21 & 28 When xy=21 A⇒x−y=4 (assume x≥y) B⇒x+y=10 x=7,y=3 (or x=3,y=7 due to symmetry) When xy=28 A⇒x−y=3 B⇒x+y=11 x=7,y=4 (or x=4,y=7 due to symmetry) (7,3),(3,7),(7,4),(4,7)

Commented bymr W last updated on 25/Jun/20

nice solution sir!

Commented bybobhans last updated on 26/Jun/20

thank you sir