Prove that $|p^2+2q^2-n|<sqrt[4]{9n}$
$begingroup$
Prove that $forall n in mathbb{N}_+,exists p,q in mathbb{Z}$ such that $|p^2+2q^2-n|<sqrt[4]{9n}$
Solution.
1) Assertion is true $forall nin[1,108]$
proof: Let $pin{0,1,2,3,4,5}$
If $nin[3p^2,3p^2+2p+1)$ :
$min(n-(2p^2+p^2),(2p^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+2p+1,3p^2+4p+2)$ :
$min(n-(2p^2+(p+1)^2),(2(p+1)^2+p^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+4p+2,3(p+1)^2]$ :
$min(n-(2(p+1)^2+p^2),(2(p+1)^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
Q.E.D.
2) Assertion is true $forall nin [108,289]$
Proof:
Assertion is true $forall nin{108,118}$
$min(n-(2.2^2+10^2),(2.3^2+10^2)-n)le 5<sqrt[4]{9times 108}lesqrt[4]{9n}$
Assertion is true $forall nin{118,129}$
$min(n-(2.3^2+10^2),(2.8^2+1^2)-n)le 5<sqrt[4]{9times 118}lesqrt[4]{9n}$
Assertion is true $forall nin{129,139}$
$min(n-(2.8^2+1^2),(2.3^2+11^2)-n)le 5<sqrt[4]{9times 129}lesqrt[4]{9n}$
Assertion is true $forall nin{139,150}$
$min(n-(2.3^2+11^2),(2.5^2+10^2)-n)le 5<sqrt[4]{9times 139}lesqrt[4]{9n}$
Assertion is true $forall nin{150,162}$
$min(n-(2.5^2+10^2),(2.3^2+12^2)-n)le 6<sqrt[4]{9times 150}lesqrt[4]{9n}$
Assertion is true $forall nin{162,172}$
$min(n-(2.3^2+12^2),(2.6^2+10^2)-n)le 5<sqrt[4]{9times 162}lesqrt[4]{9n}$
Assertion is true $forall nin{172,179}$
$min(n-(2.6^2+10^2),(2.7^2+9^2)-n)le 3<sqrt[4]{9times 172}lesqrt[4]{9n}$
Assertion is true $forall nin{179,187}$
$min(n-(2.7^2+9^2),(2.9^2+5^2)-n)le 4<sqrt[4]{9times 179}lesqrt[4]{9n}$
Assertion is true $forall nin{187,198}$
$min(n-(2.9^2+5^2),(2.1^2+14^2)-n)le 5<sqrt[4]{9times 187}lesqrt[4]{9n}$
Assertion is true $forall nin{198,209}$
$min(n-(2.1^2+14^2),(2.10^2+3^2)-n)le 5<sqrt[4]{9times 198}lesqrt[4]{9n}$
Assertion is true $forall nin{209,219}$
$min(n-(2.10^2+3^2),(2.5^2+13^2)-n)le 5<sqrt[4]{9times 209}lesqrt[4]{9n}$
Assertion is true $forall nin{219,228}$
$min(n-(2.5^2+13^2),(2.4^2+14^2)-n)le 5<sqrt[4]{9times 219}lesqrt[4]{9n}$
Assertion is true $forall nin{228,241}$
$min(n-(2.4^2+14^2),(2.6^2+13^2)-n)le 6<sqrt[4]{9times 228}lesqrt[4]{9n}$
Assertion is true $forall nin{241,251}$
$min(n-(2.6^2+13^2),(2.11^2+3^2)-n)le 5<sqrt[4]{9times 241}lesqrt[4]{9n}$
Assertion is true $forall nin{251,264}$
$min(n-(2.11^2+3^2),(2.2^2+16^2)-n)le 6<sqrt[4]{9times 251}lesqrt[4]{9n}$
Assertion is true $forall nin{264,275}$
$min(n-(2.2^2+16^2),(2.5^2+15^2)-n)le 6<sqrt[4]{9times 264}lesqrt[4]{9n}$
Assertion is true $forall nin{275,289}$
$min(n-(2.5^2+15^2),(2.12^2+15^2)-n)le 7<sqrt[4]{9times 275}lesqrt[4]{9n}$
3) Assertion is true $forall nge 290$
Proof: Let $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$ so that $nin{2p^2+1,2(p+1)^2}$
So $nin[2p^2+k^2,2p^2+(k+1)^2)$ for some $kin{1,2,...,leftlfloorsqrt{4p+3}rightrfloor}$
And so $min(n-2p^2-k^2,2p^2+(k+1)^2-n)le k$ $le leftlfloorsqrt{4p+3}rightrfloor$
And so it remains to prove that :
$leftlfloorsqrt{4p+3}rightrfloor<sqrt[4]{9n}$ where $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$
Enough to prove $sqrt{4sqrt{frac{n-1}2}+3}<sqrt[4]{9n}$, which is easily proved true $forall nge 290$
Q.E.D.
I hope someone will post a nicer solution !
number-theory inequality
asked Dec 16 '18 at 2:09
namnam
973
$endgroup$
add a comment |
$begingroup$
Prove that $forall n in mathbb{N}_+,exists p,q in mathbb{Z}$ such that $|p^2+2q^2-n|<sqrt[4]{9n}$
Solution.
1) Assertion is true $forall nin[1,108]$
proof: Let $pin{0,1,2,3,4,5}$
If $nin[3p^2,3p^2+2p+1)$ :
$min(n-(2p^2+p^2),(2p^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+2p+1,3p^2+4p+2)$ :
$min(n-(2p^2+(p+1)^2),(2(p+1)^2+p^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+4p+2,3(p+1)^2]$ :
$min(n-(2(p+1)^2+p^2),(2(p+1)^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
Q.E.D.
2) Assertion is true $forall nin [108,289]$
Proof:
Assertion is true $forall nin{108,118}$
$min(n-(2.2^2+10^2),(2.3^2+10^2)-n)le 5<sqrt[4]{9times 108}lesqrt[4]{9n}$
Assertion is true $forall nin{118,129}$
$min(n-(2.3^2+10^2),(2.8^2+1^2)-n)le 5<sqrt[4]{9times 118}lesqrt[4]{9n}$
Assertion is true $forall nin{129,139}$
$min(n-(2.8^2+1^2),(2.3^2+11^2)-n)le 5<sqrt[4]{9times 129}lesqrt[4]{9n}$
Assertion is true $forall nin{139,150}$
$min(n-(2.3^2+11^2),(2.5^2+10^2)-n)le 5<sqrt[4]{9times 139}lesqrt[4]{9n}$
Assertion is true $forall nin{150,162}$
$min(n-(2.5^2+10^2),(2.3^2+12^2)-n)le 6<sqrt[4]{9times 150}lesqrt[4]{9n}$
Assertion is true $forall nin{162,172}$
$min(n-(2.3^2+12^2),(2.6^2+10^2)-n)le 5<sqrt[4]{9times 162}lesqrt[4]{9n}$
Assertion is true $forall nin{172,179}$
$min(n-(2.6^2+10^2),(2.7^2+9^2)-n)le 3<sqrt[4]{9times 172}lesqrt[4]{9n}$
Assertion is true $forall nin{179,187}$
$min(n-(2.7^2+9^2),(2.9^2+5^2)-n)le 4<sqrt[4]{9times 179}lesqrt[4]{9n}$
Assertion is true $forall nin{187,198}$
$min(n-(2.9^2+5^2),(2.1^2+14^2)-n)le 5<sqrt[4]{9times 187}lesqrt[4]{9n}$
Assertion is true $forall nin{198,209}$
$min(n-(2.1^2+14^2),(2.10^2+3^2)-n)le 5<sqrt[4]{9times 198}lesqrt[4]{9n}$
Assertion is true $forall nin{209,219}$
$min(n-(2.10^2+3^2),(2.5^2+13^2)-n)le 5<sqrt[4]{9times 209}lesqrt[4]{9n}$
Assertion is true $forall nin{219,228}$
$min(n-(2.5^2+13^2),(2.4^2+14^2)-n)le 5<sqrt[4]{9times 219}lesqrt[4]{9n}$
Assertion is true $forall nin{228,241}$
$min(n-(2.4^2+14^2),(2.6^2+13^2)-n)le 6<sqrt[4]{9times 228}lesqrt[4]{9n}$
Assertion is true $forall nin{241,251}$
$min(n-(2.6^2+13^2),(2.11^2+3^2)-n)le 5<sqrt[4]{9times 241}lesqrt[4]{9n}$
Assertion is true $forall nin{251,264}$
$min(n-(2.11^2+3^2),(2.2^2+16^2)-n)le 6<sqrt[4]{9times 251}lesqrt[4]{9n}$
Assertion is true $forall nin{264,275}$
$min(n-(2.2^2+16^2),(2.5^2+15^2)-n)le 6<sqrt[4]{9times 264}lesqrt[4]{9n}$
Assertion is true $forall nin{275,289}$
$min(n-(2.5^2+15^2),(2.12^2+15^2)-n)le 7<sqrt[4]{9times 275}lesqrt[4]{9n}$
3) Assertion is true $forall nge 290$
Proof: Let $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$ so that $nin{2p^2+1,2(p+1)^2}$
So $nin[2p^2+k^2,2p^2+(k+1)^2)$ for some $kin{1,2,...,leftlfloorsqrt{4p+3}rightrfloor}$
And so $min(n-2p^2-k^2,2p^2+(k+1)^2-n)le k$ $le leftlfloorsqrt{4p+3}rightrfloor$
And so it remains to prove that :
$leftlfloorsqrt{4p+3}rightrfloor<sqrt[4]{9n}$ where $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$
Enough to prove $sqrt{4sqrt{frac{n-1}2}+3}<sqrt[4]{9n}$, which is easily proved true $forall nge 290$
Q.E.D.
I hope someone will post a nicer solution !
number-theory inequality
asked Dec 16 '18 at 2:09
namnam
973
$endgroup$
$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51
add a comment |
$begingroup$
Prove that $forall n in mathbb{N}_+,exists p,q in mathbb{Z}$ such that $|p^2+2q^2-n|<sqrt[4]{9n}$
Solution.
1) Assertion is true $forall nin[1,108]$
proof: Let $pin{0,1,2,3,4,5}$
If $nin[3p^2,3p^2+2p+1)$ :
$min(n-(2p^2+p^2),(2p^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+2p+1,3p^2+4p+2)$ :
$min(n-(2p^2+(p+1)^2),(2(p+1)^2+p^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+4p+2,3(p+1)^2]$ :
$min(n-(2(p+1)^2+p^2),(2(p+1)^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
Q.E.D.
2) Assertion is true $forall nin [108,289]$
Proof:
Assertion is true $forall nin{108,118}$
$min(n-(2.2^2+10^2),(2.3^2+10^2)-n)le 5<sqrt[4]{9times 108}lesqrt[4]{9n}$
Assertion is true $forall nin{118,129}$
$min(n-(2.3^2+10^2),(2.8^2+1^2)-n)le 5<sqrt[4]{9times 118}lesqrt[4]{9n}$
Assertion is true $forall nin{129,139}$
$min(n-(2.8^2+1^2),(2.3^2+11^2)-n)le 5<sqrt[4]{9times 129}lesqrt[4]{9n}$
Assertion is true $forall nin{139,150}$
$min(n-(2.3^2+11^2),(2.5^2+10^2)-n)le 5<sqrt[4]{9times 139}lesqrt[4]{9n}$
Assertion is true $forall nin{150,162}$
$min(n-(2.5^2+10^2),(2.3^2+12^2)-n)le 6<sqrt[4]{9times 150}lesqrt[4]{9n}$
Assertion is true $forall nin{162,172}$
$min(n-(2.3^2+12^2),(2.6^2+10^2)-n)le 5<sqrt[4]{9times 162}lesqrt[4]{9n}$
Assertion is true $forall nin{172,179}$
$min(n-(2.6^2+10^2),(2.7^2+9^2)-n)le 3<sqrt[4]{9times 172}lesqrt[4]{9n}$
Assertion is true $forall nin{179,187}$
$min(n-(2.7^2+9^2),(2.9^2+5^2)-n)le 4<sqrt[4]{9times 179}lesqrt[4]{9n}$
Assertion is true $forall nin{187,198}$
$min(n-(2.9^2+5^2),(2.1^2+14^2)-n)le 5<sqrt[4]{9times 187}lesqrt[4]{9n}$
Assertion is true $forall nin{198,209}$
$min(n-(2.1^2+14^2),(2.10^2+3^2)-n)le 5<sqrt[4]{9times 198}lesqrt[4]{9n}$
Assertion is true $forall nin{209,219}$
$min(n-(2.10^2+3^2),(2.5^2+13^2)-n)le 5<sqrt[4]{9times 209}lesqrt[4]{9n}$
Assertion is true $forall nin{219,228}$
$min(n-(2.5^2+13^2),(2.4^2+14^2)-n)le 5<sqrt[4]{9times 219}lesqrt[4]{9n}$
Assertion is true $forall nin{228,241}$
$min(n-(2.4^2+14^2),(2.6^2+13^2)-n)le 6<sqrt[4]{9times 228}lesqrt[4]{9n}$
Assertion is true $forall nin{241,251}$
$min(n-(2.6^2+13^2),(2.11^2+3^2)-n)le 5<sqrt[4]{9times 241}lesqrt[4]{9n}$
Assertion is true $forall nin{251,264}$
$min(n-(2.11^2+3^2),(2.2^2+16^2)-n)le 6<sqrt[4]{9times 251}lesqrt[4]{9n}$
Assertion is true $forall nin{264,275}$
$min(n-(2.2^2+16^2),(2.5^2+15^2)-n)le 6<sqrt[4]{9times 264}lesqrt[4]{9n}$
Assertion is true $forall nin{275,289}$
$min(n-(2.5^2+15^2),(2.12^2+15^2)-n)le 7<sqrt[4]{9times 275}lesqrt[4]{9n}$
3) Assertion is true $forall nge 290$
Proof: Let $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$ so that $nin{2p^2+1,2(p+1)^2}$
So $nin[2p^2+k^2,2p^2+(k+1)^2)$ for some $kin{1,2,...,leftlfloorsqrt{4p+3}rightrfloor}$
And so $min(n-2p^2-k^2,2p^2+(k+1)^2-n)le k$ $le leftlfloorsqrt{4p+3}rightrfloor$
And so it remains to prove that :
$leftlfloorsqrt{4p+3}rightrfloor<sqrt[4]{9n}$ where $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$
Enough to prove $sqrt{4sqrt{frac{n-1}2}+3}<sqrt[4]{9n}$, which is easily proved true $forall nge 290$
Q.E.D.
I hope someone will post a nicer solution !
number-theory inequality
asked Dec 16 '18 at 2:09
namnam
973
$endgroup$
Prove that $forall n in mathbb{N}_+,exists p,q in mathbb{Z}$ such that $|p^2+2q^2-n|<sqrt[4]{9n}$
Solution.
1) Assertion is true $forall nin[1,108]$
proof: Let $pin{0,1,2,3,4,5}$
If $nin[3p^2,3p^2+2p+1)$ :
$min(n-(2p^2+p^2),(2p^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+2p+1,3p^2+4p+2)$ :
$min(n-(2p^2+(p+1)^2),(2(p+1)^2+p^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
If $nin[3p^2+4p+2,3(p+1)^2]$ :
$min(n-(2(p+1)^2+p^2),(2(p+1)^2+(p+1)^2)-n)le p<sqrt[4]{27p^2}<sqrt[4]{9n}$
Q.E.D.
2) Assertion is true $forall nin [108,289]$
Proof:
Assertion is true $forall nin{108,118}$
$min(n-(2.2^2+10^2),(2.3^2+10^2)-n)le 5<sqrt[4]{9times 108}lesqrt[4]{9n}$
Assertion is true $forall nin{118,129}$
$min(n-(2.3^2+10^2),(2.8^2+1^2)-n)le 5<sqrt[4]{9times 118}lesqrt[4]{9n}$
Assertion is true $forall nin{129,139}$
$min(n-(2.8^2+1^2),(2.3^2+11^2)-n)le 5<sqrt[4]{9times 129}lesqrt[4]{9n}$
Assertion is true $forall nin{139,150}$
$min(n-(2.3^2+11^2),(2.5^2+10^2)-n)le 5<sqrt[4]{9times 139}lesqrt[4]{9n}$
Assertion is true $forall nin{150,162}$
$min(n-(2.5^2+10^2),(2.3^2+12^2)-n)le 6<sqrt[4]{9times 150}lesqrt[4]{9n}$
Assertion is true $forall nin{162,172}$
$min(n-(2.3^2+12^2),(2.6^2+10^2)-n)le 5<sqrt[4]{9times 162}lesqrt[4]{9n}$
Assertion is true $forall nin{172,179}$
$min(n-(2.6^2+10^2),(2.7^2+9^2)-n)le 3<sqrt[4]{9times 172}lesqrt[4]{9n}$
Assertion is true $forall nin{179,187}$
$min(n-(2.7^2+9^2),(2.9^2+5^2)-n)le 4<sqrt[4]{9times 179}lesqrt[4]{9n}$
Assertion is true $forall nin{187,198}$
$min(n-(2.9^2+5^2),(2.1^2+14^2)-n)le 5<sqrt[4]{9times 187}lesqrt[4]{9n}$
Assertion is true $forall nin{198,209}$
$min(n-(2.1^2+14^2),(2.10^2+3^2)-n)le 5<sqrt[4]{9times 198}lesqrt[4]{9n}$
Assertion is true $forall nin{209,219}$
$min(n-(2.10^2+3^2),(2.5^2+13^2)-n)le 5<sqrt[4]{9times 209}lesqrt[4]{9n}$
Assertion is true $forall nin{219,228}$
$min(n-(2.5^2+13^2),(2.4^2+14^2)-n)le 5<sqrt[4]{9times 219}lesqrt[4]{9n}$
Assertion is true $forall nin{228,241}$
$min(n-(2.4^2+14^2),(2.6^2+13^2)-n)le 6<sqrt[4]{9times 228}lesqrt[4]{9n}$
Assertion is true $forall nin{241,251}$
$min(n-(2.6^2+13^2),(2.11^2+3^2)-n)le 5<sqrt[4]{9times 241}lesqrt[4]{9n}$
Assertion is true $forall nin{251,264}$
$min(n-(2.11^2+3^2),(2.2^2+16^2)-n)le 6<sqrt[4]{9times 251}lesqrt[4]{9n}$
Assertion is true $forall nin{264,275}$
$min(n-(2.2^2+16^2),(2.5^2+15^2)-n)le 6<sqrt[4]{9times 264}lesqrt[4]{9n}$
Assertion is true $forall nin{275,289}$
$min(n-(2.5^2+15^2),(2.12^2+15^2)-n)le 7<sqrt[4]{9times 275}lesqrt[4]{9n}$
3) Assertion is true $forall nge 290$
Proof: Let $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$ so that $nin{2p^2+1,2(p+1)^2}$
So $nin[2p^2+k^2,2p^2+(k+1)^2)$ for some $kin{1,2,...,leftlfloorsqrt{4p+3}rightrfloor}$
And so $min(n-2p^2-k^2,2p^2+(k+1)^2-n)le k$ $le leftlfloorsqrt{4p+3}rightrfloor$
And so it remains to prove that :
$leftlfloorsqrt{4p+3}rightrfloor<sqrt[4]{9n}$ where $p=leftlfloorsqrt{frac{n-1}2}rightrfloor$
Enough to prove $sqrt{4sqrt{frac{n-1}2}+3}<sqrt[4]{9n}$, which is easily proved true $forall nge 290$
Q.E.D.
I hope someone will post a nicer solution !
number-theory inequality
number-theory inequality
asked Dec 16 '18 at 2:09
namnam
973
asked Dec 16 '18 at 2:09
namnam
973
asked Dec 16 '18 at 2:09
namnam
973
asked Dec 16 '18 at 2:09
namnam
973
asked Dec 16 '18 at 2:09
namnam
973
973
$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51
add a comment |
$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51
$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51
add a comment |
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$begingroup$
It is better to use cdot than a period for multiplication. In my first reading $2.3^2$ looked like $5.29$ (which you may write $5,29$), not $18$ You could also look at the recurrence for the Pell equation for finding $p,q$
$endgroup$
– Ross Millikan
Dec 16 '18 at 3:06
$begingroup$
compare Bambah and Chowla 1947 insa.nic.in/writereaddata/UpLoadedFiles/PINSA/…
$endgroup$
– Will Jagy
Dec 22 '18 at 18:33
$begingroup$
Do you have any solution?
$endgroup$
– nam
Dec 23 '18 at 7:51