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Alexey Ustinov
Alexey Ustinov
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If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnectedconnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems connected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

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Alexey Ustinov
Alexey Ustinov
  • 12.3k
  • 7
  • 87
  • 119

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2}\|f\|)$$R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).

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Alexey Ustinov
Alexey Ustinov
  • 12.3k
  • 7
  • 87
  • 119

If $f(x,y)$ is a "good" function, then $$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$ where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$ Usually $R_p[f]=O(p^{1/2}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$

This observation has a lot of applications in problems cnnected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).