Prove that there are no solutions in integers x and y to the equation 2x^2 +

vazelinahS

vazelinahS

Answered question

2021-11-06

Show that the equation has no solutions with integers x and y 2x2+5y2=14.

Answer & Explanation

i1ziZ

i1ziZ

Skilled2021-11-07Added 92 answers

Step 1 
To prove: The equation has no integer solutions for x or y 
2x2+5y2=14 
PROOF 
Since (a)2=a2 for all integers and since the equation contains only squares of the unknown integers, it is safe to assume that the integers x and y are nonnegative. 
The equation has no differences and the squares are never negative (or negative signs), we only need to check all integers with squares smaller than 14. The only integers with squares smaller than 14 are 0, 1, 2 and 3: 
02=0,12=1,22=4,32=9 
Thus x2 and y2 can take on the values 0, 1, 4 and 9. Let us check if there is a possible combination that makes the equation true. 
2(0)+5(0)=014 
2(0)+5(1)=514 
2(0)+5(4)=2014 
2(0)+5(9)=4514 
2(1)+5(0)=214 
2(1)+5(1)=714 
2(1)+5(4)=2214 
2(1)+5(9)=4714 
2(4)+5(0)=1614 
2(4)+5(1)=1314 
2(4)+5(4)=2814 
2(4)+5(9)=5314 
2(9)+5(0)=1814 
2(9)+5(1)=2314 
2(9)+5(4)=3814 
2(9)+5(9)=6314 
None of the possible combinations for x2 and y2 made the equation true and thus are no solution in integers x and y such that 2x2+5y2=15 
Result 
To prove it we only need to check for x and y from {0,1,2,3}, because those are only integers which squares are lesser than 14 (assuming that x,y are nonnegative without loss of generality).

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